What is 2.7: Graphs of functions?

A useful sketch records intercepts, turning points, end behaviour, symmetry and asymptotes rather than relying on a table of isolated points.

How many marks are 2.7 questions worth?

2.7 practice drills on this page are worth 2–6 marks each; marks in a real exam vary by question.

Which past exam papers has 2.7 appeared in?

We have not yet indexed a verified past-paper appearance for 2.7 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

2.7

Understand and use graphs of functions; sketch curves including polynomials, the modulus of a linear function, y = a/x and y = a/x² with asymptotes; use graph intersections to solve equations; understand proportional relationships.

Draft — not yet indexed

Graphs of functions

Worked answers and methods for 2.7 on Edexcel A-level Maths 9MA0.

Explanation

A useful sketch records intercepts, turning points, end behaviour, symmetry and asymptotes rather than relying on a table of isolated points.
For $y=a/x$ the coordinate axes are asymptotes and the relation is inverse proportionality; $y=a/x^2$ is inverse-square, symmetric about the $y$ -axis and has the same sign as $a$ .
Translations such as $y=a/(x-p)+q$ move the asymptotes to $x=p$ and $y=q$ .
The graph of $y=|f(x)|$ reflects negative outputs above the $x$ -axis; use its intersections with another graph to solve equations or inequalities.
Do not draw a reciprocal curve crossing an asymptote, and distinguish direct proportionality $y=kx$ from a relation that merely has positive correlation.

The graph of

y=4/x

: the branches approach, but never cross, the coordinate-axis asymptotes.

Worked example

Find the intersection values of $x$ for the graphs $y=|2x-3|$ and $y=x+4$ .

1.For $x\geq3/2$ , solve $2x-3=x+4$ , giving $x=7$ .
2.For $x<3/2$ , solve $-(2x-3)=x+4$ , so $-2x+3=x+4$ and $x=-1/3$ .
3.Each value lies in its required branch, so both are intersections.

Answer: $x=7$ or $x=-\dfrac13$

Common mistakes

Don't draw a reciprocal branch crossing an axis even though the axes are asymptotes.
Don't call a relation directly proportional without a graph through the origin or an equation of the form $y=kx$ .

Exam tip

On a sketch, label intercepts, turning points and asymptotes; a table of unlabelled points is not enough.

Worked practice

Tier 1 · Easy

For the curve

y=-\dfrac{6}{x}

, state both asymptotes and the two quadrants containing its branches.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	Asymptotes $x=0$ and $y=0$ ; branches in quadrants II and IV.	2
Notes
The reciprocal form has the coordinate axes as asymptotes. Since $-6/x$ is negative for $x>0$ and positive for $x<0$ , its branches lie in quadrants IV and II respectively.

(2 marks)

Tier 2 · Standard

Sketch the graph of

y=|3x+6|-2

, showing the coordinates of the vertex and all intercepts with the coordinate axes.

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	Vertex $(-2,-2)$ $x$ -intercepts $\left(-\dfrac83,0\right)$ and $\left(-\dfrac43,0\right)$ $y$ -intercept $(0,4)$	4
Notes
The vertex occurs where $3x+6=0$ , so it is $(-2,-2)$ . For the $x$ -intercepts, $\|3x+6\|=2$ , giving $3x+6=2$ or $3x+6=-2$ . Hence $x=-4/3$ or $x=-8/3$ . When $x=0$ , $y=\|6\|-2=4$ . The graph consists of two straight-line rays meeting at the vertex.

(4 marks)

Tier 3 · Hard

The curves

y=\dfrac4x

and

y=x+3

meet twice. Determine the exact coordinates of both intersections.

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$(1,4)$ and $(-4,-1)$	4
Notes
At an intersection, $4/x=x+3$ . Since $x\ne0$ , multiply by $x$ to get $x^2+3x-4=0$ . Factorising gives $(x-1)(x+4)=0$ , so $x=1$ or $x=-4$ . Substitution into $y=x+3$ gives $y=4$ or $y=-1$ .

(4 marks)

Tier 1 · Easy

y

is directly proportional to

x^3

. Given that

y=54

when

x=3

, find

y

when

x=-2

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$y=-16$	2
Notes
Write $y=kx^3$ . Using $(x,y)=(3,54)$ gives $54=27k$ , so $k=2$ . Therefore, when $x=-2$ , $y=2(-2)^3=-16$ .

(2 marks)

Tier 2 · Standard

Sketch

y=(x+2)^2(x-1)(x-4)

. Label every intercept and state the end behaviour and whether the graph crosses or touches the

x

-axis at each root.

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	The graph touches at $(-2,0)$ , crosses at $(1,0)$ and $(4,0)$ , has $y$ -intercept $(0,16)$ , and tends to $+\infty$ at both ends.	4
Notes
The roots are $-2$ , with even multiplicity, and $1$ and $4$ , each with odd multiplicity. Hence the graph touches the axis at $(-2,0)$ and crosses at $(1,0)$ and $(4,0)$ . At $x=0$ , $y=2^2(-1)(-4)=16$ . The leading term is $x^4$ , so $y\to+\infty$ as $x\to\pm\infty$ . These features pin one configuration: it remains above the axis through the touch at $-2$ , is negative only between $1$ and $4$ , and rises at both ends.

(4 marks)

Tier 3 · Hard

The curve

y=k/x^2

passes through

(2,9)

. Find

k

and hence determine the exact coordinates of every intersection of this curve with the line

y=x+1

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$k=36$ The only intersection is $(3,4)$ .	5
Notes
Using $(2,9)$ gives $9=k/4$ , so $k=36$ . At an intersection, $36/x^2=x+1$ , with $x\ne0$ , so $x^3+x^2-36=0$ . Since $x=3$ is a root, this factorises as $(x-3)(x^2+4x+12)=0$ . The quadratic factor has discriminant $16-48=-32$ , so it has no real roots. Thus $x=3$ is the only real intersection value and $y=3+1=4$ .

(5 marks)

Tier 2 · Standard

Sketch the curve

y=\dfrac{12}{x^2}-3

. Label its asymptotes and intercepts with the coordinate axes, and state its range.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	Asymptotes $x=0$ and $y=-3$ ; $x$ -intercepts $(-2,0)$ and $(2,0)$ ; no $y$ -intercept; range $y>-3$ .	4
Notes
The denominator is zero at $x=0$ , giving the vertical asymptote $x=0$ . As $\|x\|\to\infty$ , $12/x^2\to0$ from above, so the horizontal asymptote is $y=-3$ and the range is $y>-3$ . Setting $y=0$ gives $12/x^2=3$ , hence $x=\pm2$ . The curve is undefined at $x=0$ , so there is no $y$ -intercept, and its two branches are symmetric about the $y$ -axis.

(4 marks)

Tier 3 · Hard

Find every real solution of

|x-1|=\dfrac4x

, justifying the branches or domains that you reject.

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$x=\dfrac{1+\sqrt{17}}{2}$	5
Notes
The right-hand side is negative for $x<0$ , whereas the modulus is non-negative, and $x=0$ is undefined. For $0<x<1$ , the equation is $1-x=4/x$ , giving $x^2-x+4=0$ , whose discriminant is $-15$ , so there is no real root on this branch. For $x\geq1$ , the equation is $x-1=4/x$ , so $x^2-x-4=0$ . Its roots are $(1\pm\sqrt{17})/2$ , and only $(1+\sqrt{17})/2$ lies in the required branch.

(5 marks)

Tier 3 · Hard

A reciprocal curve has equation

y=\dfrac{a}{x-2}+3

. It passes through

(4,1)

. Find

a

, state both asymptotes and find the intercepts with both coordinate axes. State the position of each branch relative to the asymptotes.

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$a=-4$ Asymptotes $x=2$ and $y=3$ $x$ -intercept $\left(\dfrac{10}{3},0\right)$ and $y$ -intercept $(0,5)$ For $x>2$ , $y<3$ ; for $x<2$ , $y>3$ .	6
Notes
Using $(4,1)$ gives $1=a/(4-2)+3$ , so $a=-4$ . The translated reciprocal form has asymptotes $x=2$ and $y=3$ . For the $x$ -intercept, solve $-4/(x-2)+3=0$ , giving $3(x-2)=4$ and $x=10/3$ . At $x=0$ , $y=-4/(-2)+3=5$ . When $x>2$ , the fraction $-4/(x-2)$ is negative, so $y<3$ ; when $x<2$ , it is positive, so $y>3$ .

(6 marks)

Q10

Tier 3 · Hard

10.

A polynomial

f

has a graph that touches the

x

-axis at

x=-1

, crosses it at

x=3

, and passes through

(0,6)

. Find

f(x)

of least possible degree, state its end behaviour and solve

f(x)\geq0

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$f(x)=-2(x+1)^2(x-3)$ As $x\to-\infty$ , $f(x)\to+\infty$ ; as $x\to+\infty$ , $f(x)\to-\infty$ . $x\leq3$	6
Notes
Touching at $-1$ requires an even multiplicity, and the least possible is $2$ ; crossing at $3$ requires odd multiplicity, with least possible $1$ . Hence $f(x)=a(x+1)^2(x-3)$ . Since $f(0)=6$ , $-3a=6$ and $a=-2$ . The leading term is $-2x^3$ , giving the stated end behaviour. The squared factor is non-negative and does not change sign at $-1$ ; the sign is therefore controlled by $-2(x-3)$ . Thus $f(x)\geq0$ exactly when $x\leq3$ .

(6 marks)

Verified exam appearances

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Other points in 2 Algebra and functions

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