Learning Goal
Syllabus Requirement
Students need to understand the
graphs of functions including:

Sketching curves defined by simple equations.

Geometrical interpretation of algebraic solution of equations.

Use of intersection points of graphs of functions to solve equations.

Functions to include simple cubic functions and the reciprocal fucntion ??y = kx?? with ??x \neq 0??.

Knowledge of the term asymptote is expected.
Study Notes
In addition to sketching quadratic curves (discussed in an earlier part of the notes), you should be able to sketch simple linear functions, cubic functions, the reciprocal function and be able to make a sketch of some other function using the same techniques.
Sketching Linear Functions
To graph a linear function you just need to find two points on the line and then draw a line that passes through those two points. The two most sensible points to choose are where the line crosses each axis (if it does cross both)  these can be found by setting ??x=0?? and ??y=0??. However, any two points on the line will do, so you can pick any two values for ??x?? or ??y??.
Sketching Quadratic Functions
Every quadratic function will have a graph that is a parabola, or in simpler words, a "ushape".
Considering the
quadratic function in the form ??ax^{2} + bx + c = 0?? can help us sketch the graph.
The figure below shows the general shape of a quadratic function where ??a?? is positive:
The graph of ??y=x^{2}??.
When the ??a?? term is negative, the shape of the parabola is flipped to be upsidedown as shown in the figure below:
The graph of ??y=x^{2}??.
Next by checking the discriminant we can determine the number of real roots and hence if the curve crosses the ??x?? axis.
Calculate ??b^{2}4ac?? if:
 Positive ??\implies?? the function has two real solutions and will therefore intersect the ??x?? axis twice.
 Zero ??\implies?? then the function has a single real solution and will touch the ??x?? axis once. This will also be the turning point of the function.
 Negative ??\implies?? then the function only has complex solutions and will not cross the ??x?? axis.
Sketching Cubic Functions
There are several steps to graphing a cubic function:
 Start by noting whether the coefficient of ??x^{3}?? is positive or negative. If the coefficient is positive, then the graph of ??x^{3}?? will increase rapidly as ??x?? gets very positive and will decrease rapidly ax ??x?? gets very negative. If the coefficient is negative, then the graph of ??x^{3}?? will decrease rapidly as ??x?? gets very positive and will increase rapidly as ??x?? gets very negative. In practice this means you only have to worry about sketchning the middle part of the curve, the outer parts are defined by the coefficient of ??x^{3}??.
 Find the ??y?? intercept, by setting ??x=0?? in the cubic.
 Find the roots of the equation to find the ??x?? intercepts.
 Using all the points you have found, and the general shape of cubic graphs, shown below, sketch the chart.
 Note that you may be required to, but won't necessarily be, label the turning points of the cubic graph, which can be found by differentiating the cubic and setting the resultant expression to 0.
There are up to three different roots for a cubic function  though in cases one or more of these may coincide. The general shape of cubic charts are shown below for positive and negative coefficients of ??x^{3}??, for the various numbers of different roots.
The graph of ??y = x^{3}??. A positive coefficient of ??x^{3}?? with 1 unique root.
The graph of ??y = x^{3}??. A negative coefficient of ??x^{3}?? with 1 unique root.
The graph of ??y = x^{3} + 3x^{2}??. A positive coefficient of ??x^{3}?? with 2 unique roots.
The graph of ??y = x^{3} + 3x^{2}??. A negative coefficient of ??x^{3}?? with 2 unique roots.
The graph of ??y = x^{3}  x^{2}  x??. A positive coefficient of ??x^{3}?? with 3 unique roots.
The graph of ??y = x^{3} + x^{2} + x??. A negative coefficient of ??x^{3}?? with 3 unique roots.
Sketching Reciprocal Functions
Functions of the form ??\frac{k}{x}?? where ??k?? is a constant and ??x \neq 0?? are called
reciprocals.
Reciprocal graphs all have the same shape, which has the following characteristics:
 The graph gets closer to the ??x??axis as the value of ??x?? increases (both positively and negatively), but it never reaches the ??x?? axis, this is called the horizontal asymptote.
 The graph also never reaches the ??y??axis but approaches it as ??x?? approaches ??0??, this called the vertical asymptote.
 Graphs of this type of function are symmetrical about two lines: ??y=x?? and ??y=x??.
Below is the graph for the reciprocal function ??\frac{k}{x}??, where ??\color{Blue}{k=1}??
The graphs of ??\color{Blue}{y = \frac{1}{x}}??.
 If the value of ??k<0?? the graph is mirrored about the ??y?? axis.
Below is the graph for the reciprocal function ??\frac{k}{x}??, where ??\color{Green}{k=1}??.
The graphs of ??\color{Green}{y = \frac{1}{x}}??.
 The value of ??k?? does not change the asymptotes, but does change how quickly the graph reaches them.
Below are the graphs of two reciprocal functions of the form ??\frac{k}{x}??, for ??\color{Blue}{k=1}?? and ??\color{Red}{k=5}??:
The graphs of ??\color{Blue}{y = \frac{1}{x}}?? and ??\color{Red}{y = \frac{5}{x}}??.
Intersection Points Of Graphs To Solve Equations
An alternative approach to solving simultaneous equations is to solve them graphically. If you plot the graphs of two functions, then the point, or points, at which they cross provide a solution to both equations. The ??x?? and ??y?? coordinates can simply be read off the graph for the solution.
Worked Examples
We'll now run through an example of each of the above types of problem:
Sketching Linear Graphs
Example 1
Question
Sketch the graph of ??y = 2x + 8??.
Show Solution
Worked Solution
Let's start by identifying two points on the line, both points where the graph crosses the axes:
$$\mathrm{When} \: x = 0, \: y = 2(0) + 8$$
$$\implies y = 8$$
$$\mathrm{When} \: y = 0, \: 0 = 2x + 8$$
$$\implies x = 4$$
Now that we know the points ??(0,8)?? and ??(4,0)?? we can plot these on a graph and then draw a line through these two points.
The graph of ??y = 2x + 8??.
Sketching Cubic Graphs
Example 2
Question
Sketch the graph of ??y = x(x+3)(x+2)??.
Show Solution
Worked Solution
Let's start by noting that in the expanded version ??y = x(x+3)(x+2) = x^{3} +5x^{2} + 6x??, the coefficient of ??x^{3}?? is positive, which means the graph will rapidly go to large values of negative ??y?? on the left hand side and will rapidly go to large values of positive ??y?? on the right hand side.
The roots are easily visible as the cubic is already factorised, so we also know that the graph will cross the ??x?? axis at ??(0,0)??, ??(3,0)?? and ??(2,0)??.
We can also find the turning points by differentiating the cubic and setting to ??0??:
$$f(x) = x(x+3)(x+2)$$
$$\implies f(x) = x(x^{2} + 3x + 2x + 6)$$
$$= x(x^{2} + 5x + 6)$$
$$= x^{3} + 5x^{2} + 6x$$
Differentiating the function ??f(x)??
$$\implies f'(x) = 3x^{2} + 10x + 6$$
$$\mathrm{When} \: f'(x) = 0, \: 3x^{2} + 10x + 6 = 0$$
$$\implies x = \frac{b\pm \sqrt{b^{2}4ac}}{2a}$$
$$= \frac{10 \pm \sqrt{10^{2}  4 \times 3 \times 6} }{2 \times 3}$$
$$= \frac{10 \pm \sqrt{10072}}{6}$$
$$= \frac{10 \pm \sqrt{28}}{6}$$
$$\therefore x \approx 0.785 \: \mathrm{or} \: x \approx 2.549$$
Calculating the ??y?? values from ??f(x)?? gives ??(0.785,2.113)?? and ??(2.549,0.631)??
We can now plot the chart using the three points for the roots and the two turning points:
The graph of ??y = x(x+3)(x+2)??.
Intersection Points Of Graphs To Solve Equations
Example 3
Question
Graphically, solve the two simultaneous equations:
$$y = x + 5$$
$$y = 3x  7$$
Show Solution
Worked Solution
To solve this pair of simultaneous equations graphically we simply need to plot them. Start by finding two points on each line (where they cross the ??x?? and ??y?? axes).
For the first equation, ??y = x + 5??, this crosses the ??x?? axis at ??y = 0?? and therefore at ??(5,0)?? and crosses the ??y?? axis at ??x= 0?? and therefore at ??(0,5)??.
For the second equation, ??y = 3x  7??, this crosses the ??x?? axis at ??y = 0?? and therefore at ??\left(\frac{7}{3}, 0\right)?? and crosses the ??y?? axis at ??x=0?? and therefore at ??(0,7)??.
The graphs for both equations are plotted below and the solution to the simultaneous equations can be found at the point of intersection ??(3,2)??:
The graphs of ??y = x + 5?? (blue) and ??y = 3x  7?? (red).
Exam Questions
The table below contains every exam question that has been asked on this topic, this includes normal papers from both January
and June sittings, International papers and Specimen papers.
To see an exam question and solution simply click on the load question icon (), the question will appear below the table and
the solution can be shown by clicking the "Show Solution" button that also appears.
The full exam paper and mark schemes are also available for download by clicking on the download icons in each
row of the table ().
Load Question 
Exam Board 
Subject 
Paper 
Year 
Month 
Module 
Question No. 
Parts 
Total Marks 
Exam Paper 
Mark Scheme 

Edexcel 
Maths 
Standard 
2006 
January 
Core 1 
10 
(b) 
3 



Edexcel 
Maths 
Standard 
2006 
June 
Core 1 
3 
(a) 
3 



Edexcel 
Maths 
Standard 
2006 
June 
Core 1 
9 
(c) 
3 



Edexcel 
Maths 
Standard 
2007 
January 
Core 1 
10 
(a) 
6 



Edexcel 
Maths 
Standard 
2007 
June 
Core 1 
9 
(c) 
3 



Edexcel 
Maths 
Standard 
2008 
January 
Core 1 
10 
(a) 
4 



Edexcel 
Maths 
Standard 
2008 
June 
Core 1 
6 
(a) 
3 



Edexcel 
Maths 
Standard 
2009 
January 
Core 1 
8 
(a), (b) 
6 



Edexcel 
Maths 
Standard 
2009 
June 
Core 1 
10 
(b) 
4 



Edexcel 
Maths 
Standard 
2010 
January 
Core 1 
9 
(b) 
3 



Edexcel 
Maths 
Standard 
2010 
January 
Core 1 
10 
(c) 
3 



Edexcel 
Maths 
Standard 
2010 
June 
Core 1 
4 
(b) 
2 



Edexcel 
Maths 
Standard 
2010 
June 
Core 1 
10 
(a) 
5 



Edexcel 
Maths 
Standard 
2011 
January 
Core 1 
10 
n/a 
8 



Edexcel 
Maths 
Standard 
2011 
June 
Core 1 
10 
(a) 
4 



Edexcel 
Maths 
Standard 
2012 
January 
Core 1 
5 
(b) 
4 



Edexcel 
Maths 
Standard 
2012 
January 
Core 1 
8 
(b) 
3 



Edexcel 
Maths 
Standard 
2012 
June 
Core 1 
8 
(c) 
3 



Edexcel 
Maths 
Standard 
2013 
January 
Core 1 
10 
(b) 
4 



Edexcel 
Maths 
Standard 
2013 
June 
Core 1 
11 
(a), (b) 
3 



Edexcel 
Maths 
International 
2013 
June 
Core 1 
9 
(a) 
5 


Edexcel  Maths  Standard  2006  January  Core 1  Question 10
Question
10.
$$x^{2} + 2x + 3 \equiv (x + a)^2 + b$$
10. (b)
Sketch the graph of ??y = x^{2} + 2x + 3??, indicating clearly the coordinates of any intersections with the coordinate axes. [3]
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Worked Solution
10. (b)
From the 10.(a) we have the following:
$$ y = x^{2} + 2x + 3 \equiv (x + 1)^2 + 2$$
Recall That the function is a
quadratic therefore will have a
Ushaped curve shape.
By checking the discriminant we can determine the number of real number roots to this equation.
$$b^{2}4ac = 2^{2}  4 \times 1 \times 3 = 8$$
Since the discriminant is negative there are no real number roots to this equation; therefore will not cross the ??x?? axis.
Next calculate the point at which the curve crosses the ??y??axis, when ??x=0??:
$$y = (0 + 1)^2 + 2$$
$$y = 3$$
This gives us the coordinate for the intersection with the ??y??axis at ??(0, 3)??
We can find the minima (turning point) by differentiating and setting to ??0??:
$$y = x^{2} + 2x + 3$$
Differentiating the equation:
$$\implies \frac{\mathrm{d} y}{\mathrm{d} x} = 2x + 2$$
$$\mathrm{When} \: \frac{\mathrm{d} y}{\mathrm{d} x} = 0, \: 2x + 2 = 0$$
$$\implies 2x = 2$$
$$\implies x = 1$$
Using this in the original equation to give us the ??y?? value:
$$\implies y = (1 + 1)^2 + 2$$
$$\therefore y = 2$$
This give us the turning point coordinate of ??(1, 2)??
Using the turning point ??(1, 2)??, ??y??axis intersection at ??(0, 3)?? and knowledge that a quadratic has a
Ushaped curve with no ??x?? axis intersection, allows us to sketch the following:
The graph of ??x^{2} + 2x + 3??.
Edexcel  Maths  Standard  2006  June  Core 1  Question 3
Question
3.a
Sketch the graph of ??y = (x + 3)^2??. [3]
Show on the sketch the coordinates of each point at which the graph meets the axis.
Hide Question
Show Solution
Worked Solution
3.a
$$y = (x + 3)^2$$
Multiplying out the bracket to give:
$$x^{2} + 6x + 9$$
Recall that the function is a
quadratic therefore will have a
Ushaped curve shape.
By checking the discriminant we can determine the number of real number roots to this equation.
$$b^{2}4ac = 6^{2}  4 \times 1 \times 9 = 0$$
Since the discriminant is 0 there is one real number solution to this equation; therefore the curve will touch but not cross the ??x?? axis.
We can find the minima (turning point) by differentiating and setting to ??0??:
Differentiating the equation:
$$\implies \frac{\mathrm{d} y}{\mathrm{d} x} = 2x + 6$$
$$\mathrm{When} \: \frac{\mathrm{d} y}{\mathrm{d} x} = 0, \: 2x + 6 = 0$$
$$\implies 2x = 6$$
$$\implies x = 3$$
Using this in the original equation to give us the ??y?? value:
$$\implies y = (3 + 3)^2$$
$$\therefore y = 0$$
This give us the turning point coordinate of ??(3, 0)??
Next calculate the point at which the curve crosses the ??y??axis, when ??x=0??:
$$y = (0 + 3)^2$$
$$y = 9$$
This gives us the coordinate for the intersection with the ??y??axis at ??(0, 9)??
Using the turning point ??(3, 0)??, ??y??axis intersection at ??(0, 9) and knowledge that a quadratic has a
Ushaped curve allows us to sketch the following:
The graph of ??y = (x + 3)^2??.
Edexcel  Maths  Standard  2006  June  Core 1  Question 9
Question
9.
Given that ??f(x)=(x^{2}  6x)(x  2)+ 3x??
9.c
Sketch the graph of ??y = f(x)??, showing the coordinates of each point at which the graph meets the axis. [3]
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Worked Solution
9.c
From question sections 9(a) and 9(b) we will have fully factorised ??f(x)?? as follows:
$$f(x)=(x^{2}  6x)(x  2)+ 3x$$
$$\implies f(x)=x^{3}  8x^{2} + 15x$$
$$\implies f(x)=x(x^{2}  8x + 15)$$
$$\implies f(x)=x(x  5)(x  3)$$
Recall that the coefficient of ??x^{3}?? is positive, which means the graph will rapidly go to large values of negative ??y?? on the left hand side and will rapidly go to large values of positive ??y?? on the right hand side.
The roots are easily visible as we factorised the cubic in the first part of the question, therefore setting these to zero will give the points at which the curve crosses the ??x?? axis
$$\implies f(x)= y = x(x  5)(x  3)$$
$$\mathrm{When} \: y = 0, \: x(x  5)(x  3)=0$$
$$\implies x=0$$
$$\mathrm{Or} \: (x  5)=0$$
$$\implies x=5$$
$$\mathrm{Or} \: (x  3)=0$$
$$\implies x=3$$
So we know that the graph will cross the ??x?? axis at ??(0,0)??, ??(3,0)?? and ??(5,0)??. Since the curve passes through the origin we do not need to calculate a separate ??y?? intersept.
This is sufficient information for us to sketch the curve and indicate the points at which it meets the ??x?? axis, as required in the question.
However we could make a more accurate sketch, by also finding the turning points by differentiating the cubic and setting to ??0??:
$$f(x)=x^{3}  8x^{2} + 15x$$
Differentiating the function ??f(x)??
$$\implies f'(x) = 3x^{2}  16x + 15$$
$$\mathrm{When} \: f'(x) = 0, \: 3x^{2}  16x + 15 = 0$$
$$\implies x = \frac{b\pm \sqrt{b^{2}4ac}}{2a}$$
$$= \frac{16 \pm \sqrt{16^{2}  4 \times 3 \times 15} }{2 \times 3}$$
$$= \frac{16 \pm \sqrt{256180}}{6}$$
$$= \frac{16 \pm \sqrt{76}}{6}$$
$$\therefore x \approx 4.12 \: \mathrm{or} \: x \approx 1.21$$
Calculating the ??y?? values from ??f(x)?? gives ??(4.12,4.06)?? and ??(1.21,8.21)??
We can now plot the curve ??f(x)?? more accurately using the three points for the roots and also the two turning points:
The graph of ??y = (x^{2}  6x)(x  2)+ 3x??.
Edexcel  Maths  Standard  2007  January  Core 1  Question 10
Question
10.a
On the same axis sketch the graphs of the curves with equations:
i) ??y = x^{2}(x  2)?? [3].
ii) ??y = x(6  x)?? [3].
and indicate on your sketches the coordinates of all the points where the curves cross the ??x?? axis.
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Worked Solution
10.a
i.
From the question :
$$y = x^{2}(x  2)$$
$$\implies y=x^{3}  2x^{2}$$
Recall that the coefficient of ??x^{3}?? is positive, which means the graph will rapidly go to large values of negative ??y?? on the left hand side and will rapidly go to large values of positive ??y?? on the right hand side.
The roots are easily visible from the factorised the cubic given in the question, therefore setting these to zero will give the points at which the curve crosses the ??x?? axis
$$\implies y = x^{2}(x  2)$$
$$\mathrm{When} \: y = 0, \: x^{2}(x  2)=0$$
$$\implies x=0$$
$$\mathrm{Or} \: (x  2)=0$$
$$\implies x=2$$
So we know that the graph will cross the ??x?? axis at ??(0,0)?? and ??(2,0)??. As the curve passes through the origin this is the ??y?? axis intercept and also as there are only two roots the curve must touch the ???x?? axis at ??(2,0)??, hence is also a turning point.
This is sufficient information for us to sketch the curve and indicate the points at which it meets the ??x?? axis, as required in the question.
We could make a more accurate sketch, by finding the turning points by differentiating the cubic and setting to ??0??:
$$y=x^{3}  2x^{2}$$
Differentiating the equation:
$$\implies \frac{\mathrm{d} y}{\mathrm{d} x} = 3x^{2}  4x = x(3x  4)$$
$$\mathrm{When} \: \frac{\mathrm{d} y}{\mathrm{d} x} = 0, \: x(3x  4) = 0$$
$$\implies x=0$$
$$\mathrm{Or} \: (3x  4)=0$$
$$\implies 3x = 4$$
$$\implies x=\frac{4}{3}$$
$$\implies x = 1.33$$
Using this in the original equation to give us the ??y?? values we can say that the turning points occur at ??(0,0)?? and ??(1.33,1.185)??.
We can now plot the first curve (i) more accurately using the two points for the roots and also the two turning points.
ii.
$$y = x(6  x)$$
$$\implies y = 6x  x^{2}$$
Recall that the function is a
quadratic, with a negative ??x^{2}?? coefficient, therefore will have an inverted
Ushaped curve shape
By checking the discriminant we can determine the number of real number roots to this equation.
$$b^{2}4ac = 6^{2}  4 \times 1 \times 0 = 36$$
Since the discriminant is positive there are two real number solutions to this equation; therefore the curve will cross the ??x?? axis.
To calculate the points at which the curve crosses the ??x??axis, when ??y=0??:
$$\implies y = x(6  x)$$
$$\mathrm{When} \: y = 0, \: x(6  x)=0$$
$$\implies x=0$$
$$\mathrm{Or} \: (6  x)=0$$
$$\implies x = 6$$
This gives us the coordinate for the intersection with the ??x??axis at ??(0, 0)?? and ??(6, 0)??
We can find the turning point (in this case maximum) by differentiating and setting to ??0??:
$$y = x(6  x)$$
Multiplying out the bracket to give:
$$y = 6x  x^{2}$$
Differentiating the equation:
$$\implies \frac{\mathrm{d} y}{\mathrm{d} x} = 6  2x$$
$$\mathrm{When} \: \frac{\mathrm{d} y}{\mathrm{d} x} = 0, \: 6  2x = 0$$
$$\implies 2x = 6$$
$$\implies x = 3$$
Using this in the original equation to give us the ??y?? value:
$$\implies y = 6 \times 3  3^{2}$$
$$\implies y = 18  9$$
$$\therefore y = 9$$
This give us the turning point coordinate of ??(3, 9)??
Using the ??x??axis intersections at ??(0, 0)?? and ??(6, 0), the turning point coordinate of ??(3, 9)?? and knowledge that the quadratic has an inverted
Ushaped curve allows us to sketch both curves (i) and (ii):
The graph of ??\color{blue}{y = x^{2}(x  2)}?? and ??\color{green}{y = x(6  x)}??.
Edexcel  Maths  Standard  2007  June  Core 1  Question 9
Question
9.
The curve ??C?? with the equation ??y=f(x)?? passes through the point ??(5, 65)??.
From (b) we are given ??f(x) = x(2x + 3)(x  4)??
(c)
Sketch ??C??, showing the coordinates of the points where ??C?? crosses the ??x??axis. [3]
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Worked Solution
9.c
From question sections 9(b) we will have fully factorised ??f(x)?? as follows:
$$f(x)=x(2x + 3)(x  4)$$
$$\implies f(x)=2x^{3}  5x^{2}  12x$$
Recall that the coefficient of ??x^{3}?? is positive, which means the graph will rapidly go to large values of negative ??y?? on the left hand side and will rapidly go to large values of positive ??y?? on the right hand side.
The roots are easily visible as we have the factorised the cubic in the first part of the question, therefore setting these to zero will give the points at which the curve crosses the ??x?? axis
$$\implies f(x)= y = x(2x + 3)(x  4)$$
$$\mathrm{When} \: y = 0, \: x(2x + 3)(x  4)=0$$
$$\implies x = 0$$
$$\mathrm{Or} \: (2x + 3)=0$$
$$\implies 2x = 3$$
$$\implies x = \frac{3}{2}$$
$$\mathrm{Or} \: (x  4)=0$$
$$\implies x = 4$$
So we know that the graph will cross the ??x?? axis at ??(0,0)??, ??(\frac{3}{2},0)?? and ??(4,0)??.
This is sufficient information for us to sketch the curve and indicate the points at which it meets the ??x?? axis, as required in the question.
However we could make a more accurate sketch, by also finding the turning points by differentiating the cubic and setting to ??0??.
We are given ??f'(x)?? in the question:
$$f'(x) = 6x^{2}  10x  12$$
$$\mathrm{When} \: f'(x) = 0, \: 6x^{2}  10x  12 = 0$$
$$\implies x = \frac{b\pm \sqrt{b^{2}4ac}}{2a}$$
$$= \frac{10 \pm \sqrt{10^{2}  4 \times 6 \times 12} }{2 \times 6}$$
$$= \frac{10 \pm \sqrt{100+288}}{12}$$
$$= \frac{10 \pm \sqrt{388}}{12}$$
$$\therefore x \approx 2.47 \: \mathrm{or} \: x \approx 0.81$$
Calculating the ??y?? values from ??f(x)?? gives ??(2.47,30.01)?? and ??(0.81,5.38)??
We can now plot the curve ??f(x)?? more accurately using the three points for the roots and also the two turning points:
The graph of ??y = f(x)=x(2x + 3)(x  4)??.
Edexcel  Maths  Standard  2008  January  Core 1  Question 10
Question
10.
The curve ??C?? has equation
$$y = (x + 3)(x  1)^{2}$$
a.
Sketch ??C?? showing clearly the coordinates of the points where the curve meets the coordinate axes. [4]
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Worked Solution
10.a
From question we will have the fully factorised equation as follows:
$$y = (x + 3)(x  1)^{2}$$
Recall that the coefficient of ??x^{3}?? is positive, which means the graph will rapidly go to large values of negative ??y?? on the left hand side and will rapidly go to large values of positive ??y?? on the right hand side.
The roots are easily visible as we have the factorised the cubic given in the question, therefore setting these to zero will give the points at which the curve crosses the ??x?? axis.
Note: as there are only two roots this suggests that a root and turning point must coincide.
$$\implies y = (x + 3)(x  1)^{2}$$
$$\mathrm{When} \: y = 0, \: (x + 3)(x  1)^{2}=0$$
$$\mathrm{Or} \: (x + 3)=0$$
$$\implies x = 3$$
$$\mathrm{Or} \: (x  1)=0$$
$$\implies x = 1$$
So we know that the graph will cross the ??x?? axis at ??(3,0)?? and ??(1,0)??.
Setting ??x?? to zero will give us the point at which the curve crosses the ??y?? axis:
$$\implies y = (0 + 3)(0  1)^{2}$$
$$\implies y = 3 \times 1$$
$$\implies y = 3$$
So we know that the graph will cross the ??y?? axis at ??(0,3)??.
This is sufficient information for us to sketch the curve and indicate the points at which it meets the ??x?? and ??y?? axis, as required in the question.
However we could make a more accurate sketch, by also finding the turning points by expanding out the expression, differentiating the cubic and setting to ??0??:
$$y = (x + 3)(x  1)^{2}$$
$$\implies y = x^{3} + x^{2}  5x + 3$$
Differentiating the equation and factorising:
$$\implies \frac{\mathrm{d} y}{\mathrm{d} x} = 3x^{2} + 2x  5 = (3x + 5)(x  1)$$
$$\mathrm{When} \: \frac{\mathrm{d} y}{\mathrm{d} x} = 0, \: (3x + 5)(x  1) = 0$$
$$\implies 3x + 5 = 0$$
$$\implies 3x = 5$$
$$\implies x = \frac{5}{3}$$
$$\mathrm{Or} \: (x  1)=0$$
$$\implies x = 1$$
$$\therefore x \approx 1.66 \: \mathrm{or} \: x = 1$$
Calculating the ??y?? values from the equation gives ??(1, 0)?? and ??(1.66, 9.48)??
We can now plot the curve ??C?? more accurately using the two points for the roots and also the two turning points:
The graph of ??y = (x + 3)(x  1)^{2}??.
Edexcel  Maths  Standard  2008  June  Core 1  Question 6
Question
6.
The curve ??C?? has equation ??y = \frac{3}{x}?? and line ??l?? has equation ??y = 2x+5??.
(a) Sketch the graphs of ??C?? and ??l??, indicating clearly the coordinates of any intersections with the axes. [3]
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Worked Solution
6.
Start with the curve ??C?? with equation ??y = \frac{3}{x}??
Recall that the functions of the form ??\frac{k}{x}?? (in this case where ??k=3??) are
reciprocals.
We can sketch the graph of the reciprocal ??y = \frac{3}{x}??, noting that there is no intersection with either coordinate axes.
Next the line ??l?? with equation ??y = 2x+5??
Let's start by identifying two points on the line, both points where the graph crosses the axes:
$$\mathrm{When} \: x = 0, \: y = 2(0) + 5$$
$$\implies y = 5$$
$$\mathrm{When} \: y = 0, \: 0 = 2x + 5$$
$$\implies x = \frac{5}{2}$$
$$\implies x = 2.5$$
Now that we know the points at which the line intersects the axes ??(0,5)?? and ??(2.5,0)??, we can plot these on a graph and then draw a line through these two points.
The graph of ??\color{green}{y = 2x+5}?? & ??\color{blue}{y = \frac{3}{x}}??.
Edexcel  Maths  Standard  2009  January  Core 1  Question 8
Question
8.
The point ??P (1,a)?? lies on the curve with equation ??y = (x + 1)^{2} (2  x)??.
(a). Find the value of ??a??. [1]
(b). Sketch the curves with the following equations:
(i) ??y = (x + 1)^{2}(2  x)??
(ii) ??y = \frac{2}{x}??
On your diagram show clearly the coordinates of any points at which the curves meet the axes. [5]
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Worked Solution
8.a
As the point, ??P (1, a)?? lies on the curve with equation ??y = (x + 1)^2 (2  x)?? then we can solve by substituting ??x = 1?? and ??y = a?? in to the equation.
??\implies a = (1 + 1)^2 (2  1)??
??a = 2^2 \times 1??
??a = 4??
So the coordinates for point ??P?? are ??(1, 4)??
8.b
i) From question we will have the fully factorised equation as follows:
$$y = (x + 1)^{2}(2  x)$$
Recall that the coefficient of ??x^{3}?? is negative, which means the graph will rapidly go to large values of positive ??y?? on the left hand side and will rapidly go to large values of negative ??y?? on the right hand side.
The roots are easily visible as we have the factorised the cubic given in the question, therefore setting these to zero will give the points at which the curve crosses the ??x?? axis.
Note: as there are only two roots this suggests that a root and turning point must coincide.
$$\implies y = (x + 1)^{2}(2  x)$$
$$\mathrm{When} \: y = 0, \: (x + 1)^{2}(2  x) = 0$$
$$\mathrm{Or} \: (x + 1) = 0$$
$$\implies x = 1$$
$$\mathrm{Or} \: (2  x) = 0$$
$$\implies x = 2$$
So we know that the graph will cross the ??x?? axis at ??(1,0)?? and ??(2,0)??.
Setting ??x?? to zero will give us the point at which the curve crosses the ??y?? axis:
$$\implies y = (0 + 1)^{2}(2  0)$$
$$\implies y = 1 \times 2$$
$$\implies y = 2$$
So we know that the graph will cross the ??y?? axis at ??(0,2)??.
This is sufficient information for us to sketch the curve and indicate the points at which it meets the ??x?? and ??y?? axis, as required in the question.
However we could make a more accurate sketch, by also finding the turning points by expanding out the expression, differentiating the cubic and setting to ??0??:
$$y = (x + 1)^{2}(2  x)$$
$$\implies y = x^{3} + 3x + 2$$
Differentiating the equation and factorising:
$$\implies \frac{\mathrm{d} y}{\mathrm{d} x} = 3x^{2} + 3 = (3  3x)(x + 1)$$
$$\mathrm{When} \: \frac{\mathrm{d} y}{\mathrm{d} x} = 0, \: (3  3x)(x + 1) = 0$$
$$\implies 3  3x = 0$$
$$\implies 3x = 3$$
$$\implies x = 1$$
$$\mathrm{Or} \: (x + 1)=0$$
$$\implies x = 1$$
$$\therefore x = 1 \: \mathrm{or} \: x = 1$$
Calculating the ??y?? values from the equation gives ??(1,4)?? and ??(1,0)??
We can now plot the curve (shown in green below) more accurately using the two points ??(1,0)?? and ??(2,0)?? where the curve meets the ??x?? axis, crosses the ??y?? axis at ??(0,2)?? and also the two turning points ??(1,4)?? and ??(1,0)??.
(ii) Next taking ??y = \frac{2}{x}??
Recall that the functions of the form ??\frac{k}{x}?? (in this case where ??k = 2??) are
reciprocals.
We can sketch the graph of the reciprocal ??y = \frac{2}{x}?? (shown in blue below), noting that there is no intersection with either coordinate axes.
The graph of ??\color{green}{y = \frac{2}{x}}?? & ??\color{blue}{y = (x + 1)^{2}(2  x)}??.
Edexcel  Maths  Standard  2009  June  Core 1  Question 10
Question
10.
(b) Sketch the curve with equation
$$y = x^{3}  6x^{2} + 9x$$
showing the coordinates of the points at which the curve meets the xaxis. [4]
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Worked Solution
10.b
From question sections 10(a) we will have fully factorised the equation as follows:
$$y = x^{3}  6x^{2} + 9x$$
$$\implies y = x(x^{2}  6x + 9)$$
$$\implies y = x(x  3)(x  3)$$
$$\implies y = x(x  3)^{2}$$
Recall that the coefficient of ??x^{3}?? is positive, which means the graph will rapidly go to large values of negative ??y?? on the left hand side and will rapidly go to large values of positive ??y?? on the right hand side.
The roots are easily visible as we factorised the cubic in the first part of the question, therefore setting these to zero will give the points at which the curve crosses the ??x?? axis
$$\implies y = x(x  3)(x  3)$$
$$\mathrm{When} \: y = 0, \: x(x  3)(x  3)=0$$
$$\implies x = 0$$
$$\mathrm{Or} \: (x  3)=0$$
$$\implies x = 3$$
So we know that the graph will cross the ??x?? axis at ??(0,0)?? and ??(3,0)??. Since the curve passes through the origin we do not need to calculate a separate ??y?? axes intercept.
As there are only two roots this indicates that a root and turning point must coincide at ??(3,0)??.
This is sufficient information for us to sketch the curve and indicate the points at which it meets the ??x?? axis, as required in the question.
However we could make a more accurate sketch, by also finding the turning points by differentiating the cubic and setting to ??0??:
$$y = x^{3}  6x^{2} + 9x$$
Differentiating the equation and factorising:
$$\implies \frac{\mathrm{d} y}{\mathrm{d} x} = 3x^{2} 6x + 9 = (3x  3)(x  1)$$
$$\mathrm{When} \: \frac{\mathrm{d} y}{\mathrm{d} x} = 0, \: (3x  3)(x  1) = 0$$
$$\implies 3x  3 = 0$$
$$\implies 3x = 3$$
$$\implies x = 1$$
$$\mathrm{Or} \: (x  1)=0$$
$$\implies x = 1$$
$$\therefore x = 3 \: \mathrm{or} \: x = 1$$
Calculating the ??y?? values from the equation gives ??(1,4)?? and ??(3,0)??
We can now plot the curve more accurately using the two points ??(0,0)?? and ??(3,0)?? where the curve meets the ??x?? axis and also the two turning points ??(1,4)?? and ??(3,0)??.
The graph of ??y = x^{3}  6x^{2} + 9x??.
Edexcel  Maths  Standard  2010  January  Core 1  Question 9
Question
9.
(b). Sketch the curve ??C?? with equation:
$$y = x^{3}  4x$$
showing the coordinates of the points at which the curve meets the xaxis. [3]
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Worked Solution
9.b
From question sections 9(a) we will have fully factorised the equation as follows:
$$y = x^{3}  4x$$
$$\implies y = x(x^{2}  4)$$
$$\implies y = x(x  2)(x + 2)$$
Recall that the coefficient of ??x^{3}?? is positive, which means the graph will rapidly go to large values of negative ??y?? on the left hand side and will rapidly go to large values of positive ??y?? on the right hand side.
The roots are easily visible as we factorised the cubic in the first part of the question, therefore setting these to zero will give the points at which the curve crosses the ??x?? axis
$$\implies y = x(x  2)(x + 2)$$
$$\mathrm{When} \: y = 0, \: x(x  2)(x + 2)=0$$
$$\implies x = 0$$
$$\mathrm{Or} \: (x  2)=0$$
$$\implies x = 2$$
$$\mathrm{Or} \: (x + 2)=0$$
$$\implies x = 2$$
So we know that the graph will cross the ??x?? axis at ??(0,0)??, ??(2,0)?? and ??(2,0)??. Since the curve passes through the origin we do not need to calculate a separate ??y?? axes intercept.
This is sufficient information for us to sketch the curve and indicate the points at which it meets the ??x?? axis, as required in the question.
However we could make a more accurate sketch, by also finding the turning points by differentiating the cubic and setting to ??0??:
$$y = x^{3}  4x$$
Differentiating the equation and solving:
$$\implies \frac{\mathrm{d} y}{\mathrm{d} x} = 3x^{2}  4$$
$$\mathrm{When} \: \frac{\mathrm{d} y}{\mathrm{d} x} = 0, \: 3x^{2}  4 = 0$$
$$\implies x = \frac{b\pm \sqrt{b^{2}4ac}}{2a}$$
$$= \frac{0 \pm \sqrt{0^{2}  4 \times 3 \times 4} }{2 \times 3}$$
$$= \frac{0 \pm \sqrt{48}}{6}$$
$$\therefore x \approx 1.15 \: \mathrm{or} \: x \approx 1.15$$
Calculating the ??y?? values from the equation gives ??(1.15,3.08)?? and ??(1.15,3.08)??
We can now plot the curve more accurately using the points ??(0,0)??, ??(2,0)?? and ??(2,0)?? where the curve meets the ??x?? axis and also the two turning points ??(1.15,3.08)?? and ??(1.15,3.08)??.
The graph of ??y = x^{3}  4x??.
Edexcel  Maths  Standard  2010  January  Core 1  Question 10
Question
10.c
??f(x)=x^{2} + 4kx + (3+11k)??, where ??k?? is a constant.
The equation ??f(x) = 0?? has no real roots.
Given that ??k=1??, sketch the graph of ??y=f(x)??, showing the coordinates of any point at which the graph crosses a coordinate axis. [3]
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Worked Solution
10.c
Start by substituting ??k=1?? into the given equation:
$$f(x)= y = x^{2} + 4kx + (3+11k)$$
$$\implies y = x^{2} + (4 \times 1)x +(3 + 11 \times 1)$$
$$\implies y = x^{2} + 4x + 14$$
Recall that the function is a
quadratic therefore will have a
Ushaped curve shape.
We are told in the question that the equation ??f(x) = 0?? has no real roots, therefore the curve will not cross the ??x?? axis. ie the discriminant is negative.
Start by calculating the point at which the curve crosses the ??y??axis, when ??x=0??:
$$\implies y = 0^{2} + (4 \times 0) + 14$$
$$\therefore y = 14$$
This gives us the coordinate for the intersection with the ??y??axis at ??(0, 14)??
We can find the turning point by differentiating and setting to ??0??:
$$y = x^{2} + 4x + 14$$
$$\implies \frac{\mathrm{d} y}{\mathrm{d} x} = 2x + 4$$
$$\mathrm{When} \: \frac{\mathrm{d} y}{\mathrm{d} x} = 0, \: 2x + 4 = 0$$
$$\implies 2x = 4$$
$$\implies x = 2$$
Using this in the original equation to give us the ??y?? value:
$$\implies y = (2)^2 + (4 \times 2) + 14$$
$$\therefore y = 10$$
This give us the turning point coordinate of ??(2, 10)??
Using the turning point ??(2, 10)??, ??y??axis intersection at ??(0, 14)?? and knowledge that a quadratic has a
Ushaped curve allows us to sketch the following:
The graph of ??y = x^{2} + 4x + 14??.
Edexcel  Maths  Standard  2010  June  Core 1  Question 4
Question
4.b
Sketch the curve with equation ??y = x^{2} + 6x + 11??, showing clearly any intersections with the coordinate axes. [2]
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Worked Solution
4.b
$$y = x^{2} + 6x + 11$$
Recall that the function is a
quadratic therefore will have a
Ushaped curve shape.
By checking the discriminant we can determine the number of real number roots to this equation.
$$b^{2}4ac = 6^{2}  4 \times 1 \times 11 = 36  44 = 8$$
Since the discriminant is negative there are no real number roots to this equation; therefore will not cross the ??x?? axis.
We can calculate the point at which the curve crosses the ??y??axis, when ??x=0??:
$$\implies y = 0^{2} + (6 \times 0) + 11$$
$$\therefore y = 11$$
This gives us the coordinate for the intersection with the ??y??axis at ??(0, 11)??
We can find the turning point by differentiating and setting to ??0??:
$$y = x^{2} + 6x + 11$$
$$\implies \frac{\mathrm{d} y}{\mathrm{d} x} = 2x + 6$$
$$\mathrm{When} \: \frac{\mathrm{d} y}{\mathrm{d} x} = 0, \: 2x + 6 = 0$$
$$\implies 2x = 6$$
$$\implies x = 3$$
Using this in the original equation to give us the ??y?? value:
$$\implies y = (3)^2 + (6 \times 3) + 11$$
$$\therefore y = 2$$
This give us the turning point coordinate of ??(3, 2)??
Using the turning point ??(3, 2)??, ??y??axis intersection at ??(0, 11)?? and knowledge that a quadratic has a
Ushaped curve that does not cross the ??x?? axis allows us to sketch the following:
The graph of ??y = x^{2} + 6x + 11??.
Edexcel  Maths  Standard  2010  June  Core 1  Question 10
Question
10a.
Sketch the graphs of:
i) ??y = x(4  x)??
ii) ??y = x^{2}(7 – x)??
Showing clearly the coordinates of the points where the curves cross the coordinate axes. [5]
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Worked Solution
10.
i.
??y = x(4  x)??
$$\implies y = 4x  x^{2}$$
Recall that the function is a
quadratic, with a negative ??x^{2}?? coefficient, therefore will have an inverted
Ushaped curve shape
By checking the discriminant we can determine the number of real number roots to this equation.
$$b^{2}4ac = 4^{2}  4 \times 1 \times 0 = 16$$
Since the discriminant is positive there are two real number solutions to this equation; therefore the curve will cross the ??x?? axis.
Calculate the points at which the curve crosses the ??x??axis, when ??y=0??:
$$\implies y = x(4  x)$$
$$\mathrm{When} \: y = 0, \: x(4  x)=0$$
$$\implies x=0$$
$$\mathrm{Or} \: (4  x)=0$$
$$\implies x = 4$$
This gives us the coordinate for the intersection with the ??x??axis at ??(0, 0)?? and ??(4, 0)??
We can find the turning point (in this case maximum) by differentiating and setting to ??0??:
$$y = 4x  x^{2}$$
Differentiating the equation:
$$\implies \frac{\mathrm{d} y}{\mathrm{d} x} = 4  2x$$
$$\mathrm{When} \: \frac{\mathrm{d} y}{\mathrm{d} x} = 0, \: 4  2x = 0$$
$$\implies 2x = 4$$
$$\implies x = 2$$
Using this in the original equation to give us the ??y?? value:
$$\implies y = 4 \times 2  2^{2}$$
$$\implies y = 8  4$$
$$\therefore y = 4$$
This give us the turning point coordinate of ??(2, 4)??
Using these two points and knowledge that the quadratic has an inverted
Ushaped curve allows us to sketch the curve of ??y = x(4  x)?? shown below in blue.
ii.
From the question :
$$y = x^{2}(7  x)$$
$$\implies y=7x^{2}  x^{3}$$
Recall that the coefficient of ??x^{3}?? is negative, which means the graph will rapidly go to large values of positive ??y?? on the left hand side and will rapidly go to large values of negative ??y?? on the right hand side.
The roots are easily visible from the factorised cubic given in the question, therefore setting these to zero will give the points at which the curve crosses the ??x?? axis
$$\implies y = x^{2}(7  x)$$
$$\mathrm{When} \: y = 0, \: x^{2}(7  x)=0$$
$$\implies x=0$$
$$\mathrm{Or} \: (7  x)=0$$
$$\implies x=7$$
So we know that the graph will cross the ??x?? axis at ??(0,0)?? and ??(7,0)??. As the curve passes through the origin this is the ??y?? axis intercept. As there are only two roots the curve must touch the ??x?? axis at the origin, hence this is also a turning point.
This is sufficient information for us to sketch the curve and indicate the points at which it meets the coordinate axes, as required in the question.
We could make a more accurate sketch, by finding the turning points by differentiating the cubic and setting to ??0??:
$$y = 7x^{2}  x^{3}$$
Differentiating the equation:
$$\implies \frac{\mathrm{d} y}{\mathrm{d} x} = 14x  3x^{2} = x(14  3x)$$
$$\mathrm{When} \: \frac{\mathrm{d} y}{\mathrm{d} x} = 0, \: x(14  3x) = 0$$
$$\implies x=0$$
$$\mathrm{Or} \: (14  3x)=0$$
$$\implies 3x = 14$$
$$\implies x=\frac{14}{3}$$
$$\implies x = 4.67$$
Using this in the original equation to give us the ??y?? values we can say that the turning points occur at ??(0,0)?? and ??(4.67,50.81)??.
We can now plot the curve (ii) of ??y = x^{2}(7 – x)?? more accurately using the two points where it crosses the ??x?? axis at ??(0,0)?? and ??(7,0)?? and also the two turning points, ??(0,0)?? and ??(4.67,50.81)??.
The graph of ??\color{blue}{y = x(4  x)}?? and ??\color{green}{y = x^{2}(7 – x)}??.
Edexcel  Maths  Standard  2011  January  Core 1  Question 10
Question
10.
a.
Sketch the graphs of:
i) ??y = x(x+2)(3x)??
ii) ??y =  \frac{2}{x}??
Showing clearly the coordinates of the points where the curves cross the coordinate axes. [6]
b.
Using your sketch state, giving a reason, the number of real solutions to the equation. [2]
$$x(x+2)(3x)+ \frac{2}{x} = 0$$
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Worked Solution
10.a
i.
From question we will have the fully factorised equation as follows:
$$y = x(x + 2)(3  x)$$
Recall that the coefficient of ??x^{3}?? is negative, which means the graph will rapidly go to large values of positive ??y?? on the left hand side and will rapidly go to large values of negative ??y?? on the right hand side.
The roots are easily visible as we have the factorised the cubic given in the question, therefore setting these to zero will give the points at which the curve crosses the ??x?? axis.
$$\implies y = x(x + 2)(3  x)$$
$$\mathrm{When} \: y = 0, \: x(x + 2)(3  x) = 0$$
$$\implies x = 0$$
$$\mathrm{Or} \: (x + 2) = 0$$
$$\implies x = 2$$
$$\mathrm{Or} \: (3  x) = 0$$
$$\implies x = 3$$
So we know that the graph will cross the ??x?? axis at ??(2,0)??, ??(3,0)?? and pass through the origin ??(0,0)??.
This is sufficient information for us to sketch the curve and indicate the points at which it meets the ??x?? axis, as required in the question.
However we could make a more accurate sketch, by also finding the turning points by expanding out the expression, differentiating the cubic and setting to ??0??:
$$y = x(x + 2)(3  x)$$
$$\implies y = x^{3} + x^{2} + 6x$$
Differentiating the equation and solving:
$$\implies \frac{\mathrm{d} y}{\mathrm{d} x} = 3x^{2} + 2x + 6$$
$$\mathrm{When} \: \frac{\mathrm{d} y}{\mathrm{d} x} = 0, \: 3x^{2} + 2x + 6 = 0$$
$$\implies x = \frac{b\pm \sqrt{b^{2}4ac}}{2a}$$
$$= \frac{2 \pm \sqrt{2^{2}  4 \times 3 \times 6} }{2 \times 3}$$
$$= \frac{2 \pm \sqrt{76}}{6}$$
$$\therefore x \approx 1.12 \: \mathrm{or} \: x \approx 1.79$$
Calculating the ??y?? values from the equation gives ??(1.12,4.06)?? and ??(1.79,8.21)??
We can now plot the curve more accurately using the three points ??(2,0)??, ??(0,0)?? and ??(3,0)?? where the curve meets the ??x?? axis and also the two turning points ??(1.12,4.06)?? and ??(1.79,8.21)??.
ii.
Next taking ??y =  \frac{2}{x}??
Recall that the functions of the form ??\frac{k}{x}?? (in this case where ??k = 2??) are
reciprocals. Take care to sketch in the correct quadrants since the value of ??k?? is negative.
We can sketch the graph of the reciprocal ??y = \frac{2}{x}??, noting that there is no intersection with either coordinate axes.
The graph of ??\color{green}{y = \frac{2}{x}}?? & ??\color{blue}{y = x(x + 2)(3  x)}??.
b.
We are given:
$$x(x+2)(3x)+ \frac{2}{x} = 0$$
$$\implies x(x+2)(3x) =  \frac{2}{x}$$
Which will be true when the two curves of ??x(x+2)(3x)?? and ?? \frac{2}{x}?? intercept.
We can see from the sketch that there are
two intersections therefore there are
two real solutions to the equation given.
Edexcel  Maths  Standard  2011  June  Core 1  Question 10
Question
10.a
The curve ??C?? has equation
$$y = (x + 1)(x + 3)^{2}$$
Sketch ??C??, showing the coordinates of the points at which ??C?? meets the axes. [4]
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Worked Solution
10.a
From question we will have the fully factorised equation as follows:
$$y = (x + 1)(x + 3)^{2}$$
Recall that the coefficient of ??x^{3}?? is positive, which means the graph will rapidly go to large values of negative ??y?? on the left hand side and will rapidly go to large values of positive ??y?? on the right hand side.
The roots are easily visible as we have the factorised the cubic given in the question, therefore setting these to zero will give the points at which the curve crosses the ??x?? axis.
Note: as there are only two roots this suggests that a root and turning point must coincide.
$$\implies y = (x + 1)(x + 3)^{2}$$
$$\mathrm{When} \: y = 0, \: (x + 1)(x + 3)^{2}=0$$
$$\mathrm{Or} \: (x + 1)=0$$
$$\implies x = 1$$
$$\mathrm{Or} \: (x + 3)=0$$
$$\implies x = 3$$
So we know that the graph will cross the ??x?? axis at ??(3,0)?? and ??(1,0)??.
Setting ??x?? to zero will give us the point at which the curve crosses the ??y?? axis:
$$\implies y = (0 + 1)(0 + 3)^{2}$$
$$\implies y = 1 \times 9$$
$$\implies y = 9$$
So we know that the graph will cross the ??y?? axis at ??(0,9)??.
This is sufficient information for us to sketch the curve and indicate the points at which it meets the ??x?? and ??y?? axis, as required in the question.
However we could make a more accurate sketch, by also finding the turning points by expanding out the expression, differentiating the cubic and setting to ??0??:
$$y = (x + 1)(x + 3)^{2}$$
$$\implies y = x^{3} + 7x^{2} + 15x + 9$$
Differentiating the equation and factorising:
$$\implies \frac{\mathrm{d} y}{\mathrm{d} x} = 3x^{2} + 14x + 15 = (3x + 5)(x + 3)$$
$$\mathrm{When} \: \frac{\mathrm{d} y}{\mathrm{d} x} = 0, \: (3x + 5)(x + 3) = 0$$
$$\implies 3x + 5 = 0$$
$$\implies 3x = 5$$
$$\implies x = \frac{5}{3}$$
$$\mathrm{Or} \: (x + 3)=0$$
$$\implies x = 3$$
$$\therefore x \approx 1.66 \: \mathrm{or} \: x = 3$$
Calculating the ??y?? values from the equation gives ??(3, 0)?? and ??(1.66, 1.18)??
We can now plot the curve ??C?? more accurately using the two points where the graph will cross the ??x?? axis ??(3,0)?? and ??(1,0)??, the ??y?? axis intercept at ??(0,9)?? and also the turning points ??(3, 0)?? and ??(1.66, 1.18)??.
The graph of ??y = (x + 1)(x + 3)^{2}??.
Edexcel  Maths  Standard  2012  January  Core 1  Question 5
Question
5.b
The curve ??C?? has equation ??y = x(5 – x)?? and the line ??L?? has equation ??2y = 5x +4??
Sketch ??C?? and ??L?? on the same diagram, showing the coordinates of the points at which ??C?? and ??L?? meet the axes. [4]
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Worked Solution
5.b
$$y = x(5  x)$$
$$\implies y = 5x  x^{2}$$
Recall that the function is a
quadratic, with a negative ??x^{2}?? coefficient, therefore will have an inverted
Ushaped curve shape
By checking the discriminant we can determine the number of real number roots to this equation.
$$b^{2}4ac = 5^{2}  4 \times 1 \times 0 = 25$$
Since the discriminant is positive there are two real number solutions to this equation; therefore the curve will cross the ??x?? axis.
First calculate the points at which the curve crosses the ??x??axis, when ??y=0??:
$$\implies y = x(5  x)$$
$$\mathrm{When} \: y = 0, \: x(5  x)=0$$
$$\implies x=0$$
$$\mathrm{Or} \: (5  x)=0$$
$$\implies x = 5$$
This gives us the coordinate for the intersection with the ??x??axis at ??(0, 0)?? and ??(5, 0)??
We can find the turning point (in this case maximum) by differentiating and setting to ??0??:
$$y = x(5  x)$$
Multiplying out the bracket to give:
$$y = 5x  x^{2}$$
Differentiating the equation:
$$\implies \frac{\mathrm{d} y}{\mathrm{d} x} = 5  2x$$
$$\mathrm{When} \: \frac{\mathrm{d} y}{\mathrm{d} x} = 0, \: 5  2x = 0$$
$$\implies 2x = 5$$
$$\implies x = \frac{5}{2}$$
$$\implies x = 2.5$$
Using this in the original equation to give us the ??y?? value:
$$\implies y = 5 \times 2.5  2.5^{2}$$
$$\implies y = 12.5  6.25$$
$$\therefore y = 6.25$$
This give us the turning point coordinate of ??(2.5, 6.25)??
Using these two points and knowledge that the quadratic has an inverted
Ushaped curve allows us to sketch the curve ??C??.
Now considering the line ??L?? with equation ??2y = 5x +4??
Let's start by identifying two points on the line, both points where the graph crosses the axes:
$$\mathrm{When} \: x = 0, \: 2y = 2(0) + 4$$
$$\implies y = 2$$
$$\mathrm{When} \: y = 0, \: 0 = 5x + 4$$
$$\implies x = \frac{4}{5}$$
$$\implies x = 0.8$$
Now that we know the points ??(0,2)?? and ??(0.8,0)?? we can plot these on a graph and then draw a line through these two points.
The graph of ??\color{blue}{y = x(5  x)}?? and ??\color{red}{2y = 5x +4}??.
Edexcel  Maths  Standard  2012  January  Core 1  Question 8
Question
8.b
The curve ?? C_{1}?? has equation
$$y = x^{2}(x + 2)$$
Sketch ??C_{1}??, showing the coordinates of the points where ??C_{1}?? meets the ??x??axis. [3]
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Worked Solution
8.b
From question we will have the fully factorised equation as follows:
$$y = x^{2}(x + 2)$$
Recall that the coefficient of ??x^{3}?? is positive, which means the graph will rapidly go to large values of negative ??y?? on the left hand side and will rapidly go to large values of positive ??y?? on the right hand side.
The roots are easily visible as we have the factorised the cubic given in the question, therefore setting these to zero will give the points at which the curve crosses the ??x?? axis.
Note: as there are only two roots this suggests that a root and turning point must coincide.
$$\implies y = x^{2}(x + 2)$$
$$\mathrm{When} \: y = 0, \: x^{2}(x + 2)=0$$
$$\mathrm{Or} \: (x + 2)=0$$
$$\implies x = 2$$
$$\mathrm{Or} \: x^{2}=0$$
$$\implies x = 0$$
So we know that the graph will cross the ??x?? axis at ??(2,0)?? and pass through the origin ??(0,0)??
This is sufficient information for us to sketch the curve and indicate the points at which it meets the ??x?? axis, as required in the question.
However we could make a more accurate sketch, by also finding the turning points by expanding out the expression, differentiating the cubic and setting to ??0??:
$$y = x^{2}(x + 2)$$
$$\implies y = x^{3} + 2x^{2}$$
Differentiating the equation and factorising:
$$\implies \frac{\mathrm{d} y}{\mathrm{d} x} = 3x^{2} + 4x = x(3x + 4)$$
$$\mathrm{When} \: \frac{\mathrm{d} y}{\mathrm{d} x} = 0, \: x(3x + 4) = 0$$
$$\implies 3x + 4 = 0$$
$$\implies 3x = 4$$
$$\implies x = \frac{4}{3}$$
$$\mathrm{Or} \: x=0$$
$$\therefore x \approx 1.33 \: \mathrm{or} \: x = 0$$
Calculating the ??y?? values from the equation gives ??(1.33, 1.185)?? and ??(0,0)??
We can now plot the curve ??C_{1}?? more accurately using the two points where the graph will cross the ??x?? axis ??(2,0)?? and ??(0,0)?? and also the turning points ??(1.33, 1.185)?? and ??(0,0)??.
The graph of ??y = x^{2}(x + 2)??.
Edexcel  Maths  Standard  2012  June  Core 1  Question 8
Question
8.c
Sketch the curve with the equation
$$y = 4x  5  x^{2}$$
Showing clearly the coordinates of any points where the curve crosses the coordinate axis. [3]
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8.c
$$y = 4x  5  x^{2}$$
Recall that the function is a
quadratic, with a negative ??x^{2}?? coefficient, therefore will have an inverted
Ushaped curve shape
From section 8(b) we know that the discriminant is negative so has no real number roots and therefore will not cross the ??x?? axis
To calculate the point at which the curve crosses the ??y??axis, set ??x=0??:
$$y = 4(0)  5  (0)^{2}$$
$$y = 5$$
This gives us the coordinate for the intersection with the ??y??axis at ??(0, 5)??
This is sufficient information for us to sketch the curve and indicate the point at which it meets the ??y?? axis, as required in the question.
We could however make a more accurate sketch by finding the turning point (in this case maximum as negative ??x^{2}?? coefficient) by differentiating and setting to ??0??:
$$y = 4x  5  x^{2}$$
Differentiating the equation:
$$\implies \frac{\mathrm{d} y}{\mathrm{d} x} = 4  2x$$
$$\mathrm{When} \: \frac{\mathrm{d} y}{\mathrm{d} x} = 0, \: 4  2x = 0$$
$$\implies 2x = 4$$
$$\implies x = 2$$
Using this in the original equation to give us the ??y?? value:
$$\implies y = 4 \times 2  5  2^{2}$$
$$\therefore y = 1$$
This give us the turning point coordinate of ??(2, 1)??
Using the ??y?? axis intersection, turning point and knowledge that the quadratic has an inverted
Ushaped curve which does not cross the ??x?? axis allows us to sketch the curve:
The graph of ??y = 4x  5  x^{2}??.
Edexcel  Maths  Standard  2013  January  Core 1  Question 10
Question
10.b
Sketch the curve with equation ??y = 4x^{2} + 8x + 3??, showing clearly the coordinates of any points where the curve crosses the coordinate axes. [4]
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10.b
$$y = 4x^{2} + 8x + 3$$
Recall that the function is a
quadratic therefore will have a
Ushaped curve shape.
By checking the discriminant we can determine the number of real number roots to this equation.
$$b^{2}4ac = 8^{2}  4 \times 4 \times 3 = 64  48 = 16$$
Since the discriminant is positive there are two real number solutions to this equation; therefore the curve will cross the ??x?? axis.
First calculate the points at which the curve crosses the ??x??axis, when ??y=0??:
$$y = 4x^{2} + 8x + 3$$
$$\implies y = (2x + 3)(2x + 1)$$
$$\mathrm{When} \: y = 0, \: (2x + 3)(2x + 1) = 0$$
$$\implies(2x + 3) = 0$$
$$\implies x = \frac{2}{3}$$
$$\mathrm{Or} \: (2x + 1) = 0$$
$$\implies x = \frac{1}{2}$$
This gives us the coordinates for the intersection with the ??x??axis at ??(\frac{2}{3}, 0)?? and ??(\frac{1}{2}, 0)??
We can find the turning point (in this case maximum) by differentiating and setting to ??0??:
$$y = 4x^{2} + 8x + 3$$
Differentiating the equation:
$$\implies \frac{\mathrm{d} y}{\mathrm{d} x} = 8x + 8$$
$$\mathrm{When} \: \frac{\mathrm{d} y}{\mathrm{d} x} = 0, \: 8x + 8 = 0$$
$$\implies 8x = 8$$
$$\implies x = 1$$
Using this in the original equation to give us the ??y?? value:
$$\implies y = 4 \times 1^{2} + 8 \times 1 + 3$$
$$\implies y = 4  8 + 3$$
$$\therefore y = 1$$
This give us the turning point coordinate of ??(1, 1)??
Next calculate the point at which the curve crosses the ??y??axis, when ??x=0??:
$$\implies y = 4 \times 0^{2} + (8 \times 0) + 3$$
$$\therefore y = 3$$
This gives us the coordinate for the intersection with the ??y??axis at ??(0, 3)??
Using the ??x??axis intersections at ??(\frac{2}{3}, 0)?? and ??(\frac{1}{2}, 0)??, the turning point coordinate of ??(1, 1)??, ??y??axis intersection at ??(0, 3)?? and knowledge that the quadratic has an inverted
Ushaped curve allows us to sketch the curve ??C??.
The graph of ??y = 4x^{2} + 8x + 3??.
Edexcel  Maths  Standard  2013  June  Core 1  Question 11
Question
11.
Figure 2.
Figure 2 shows a sketch of the curve ??H?? with equation ??y = \frac{3}{x} + 4, x \neq{0}??
a.
Give the coordinates of the point where ??H?? crosses the ??x??axis. [1]
b.
Give the equations of the asymptotes to ??H??. [2]
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11.
a.
Where ??H?? crosses the ??x??axis, ??y = 0??
$$\implies \frac{3}{x} + 4 = 0$$
$$\implies \frac{3}{x} = 4$$
$$\implies x = \frac{3}{4}$$
Therefore the coordinate where ??H?? crosses the ??x??axis is ??(\frac{3}{4},0)??
b.
Recall that the horizontal asymptote for a reciprocal of form ??\frac{k}{x}?? is ??y=0??
Therefore the horizontal asymptote of ??H??, is ??y = 0 + 4 = 4??
Recall that the vertical asymptote for a reciprocal of form ??\frac{k}{x}?? is ??x=0??
This will remain unchanged therefore we can say the vertical asymptote of ??H??, is ??x=0??
??\frac{3}{x} + 4?? showing the vertical and horizontal asymptotes.
Edexcel  Maths  International  2013  June  Core 1  Question 9
Question
9.
Figure 1.
Figure 1 shows a sketch of the curve ??C?? with equation ??y = f(x)??
The curve ??C?? passes through the point ??(1, 0)?? and touches the ??x??axis at the point ??(2, 0)??
The curve has a maximum at the point ??(0, 4)??
a.
The equation of the curve ??C?? can be written in the form
$$y = x^{3} + ax^{2} + bx + c$$
where ??a, b?? and ??c?? are integers.
Calculate the values of ??a, b?? and ??c??. [5]
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9.a
There are several methods to obtaining the solution to this question but lets approach it in the same way we have sketched cubic functions, only in reverse!
Firstly we are given the ??y?? axis intercept, when ??x=0?? at ??(0, 4)??. Therefore substituting these values into ??y = x^{3} + ax^{2} + bx + c??:
$$4 = (0)^{3} + a(0)^{2} + b(0) + c$$
$$\implies c = 4$$
Next consider the two turning points ??(0,4)?? and ??(2,0)??. We know at these points that ??\frac{\mathrm{d} y}{\mathrm{d} x} = 0??
Differentiating the equation:
$$y = x^{3} + ax^{2} + bx + c$$
$$\implies \frac{\mathrm{d} y}{\mathrm{d} x} = 3x^{2} + 2ax + b$$
At each turning point ??\frac{\mathrm{d} y}{\mathrm{d} x} = 0, \implies 3x^{2} + 2ax + b = 0??
Considering the turning point ??(0,4)??:
$$x = 0$$
$$\implies 3(0)^{2} + 2(0)+ b = 0$$
$$\implies b = 0$$
Now applying ??b=0?? and considering the turning point ??(2,0)??:
$$x = 2$$
$$\implies 3(2)^{2} + 2(2)a + 0 = 0$$
$$\implies 12 + 4a = 0$$
$$\implies 4a = 12$$
$$\implies a = 3$$
Therefore we can say ??a = 3??, ??b = 0?? and ??c = 4??. For completeness substitute these into the equation ??y = x^{3} + ax^{2} + bx + c?? :
$$y = x^{3}  3x^{2} + 4$$
Exam Tips

When attempting to sketch a graph of a function take the following general approach:
Recognise what type of function the question is referring to, either linear, quadratic, cubic or reciprocal
Learn and recall the general shape of the function you are attempting to sketch
Attempt to find any locations at which the curve crosses either axis
For a more accurate sketch calculate the turning points however this is rarely specified in the question
You should clearly state the intercepts in your solutions and check you have provided all of the points in the question

The majority of exam questions will ask for the ??x?? axis intercepts therefore practice factorising equations so you can quickly find the roots

When sketching a quadratic checking the discriminant (??b^{2}4ac??) will help you determine if and how many times the curve crosses the ??x?? axis.
Positive ??\implies?? the function has two real solutions and will therefore intersect the ??x?? axis twice.
Zero ??\implies?? then the function has a single real solution and will touch the ??x?? axis once. This will also be the turning point of the function.
Negative ??\implies?? then the function only has complex solutions and will not cross the ??x?? axis.