Standard Level · Functions

Linear and Quadratic Functions

Linear equations, gradients and straight-line graphs, Key Features, Forms of equations, Quadratic Formula, Discriminant, Equations, Inequalities

DifficultyAll difficulties
Exam modePaper 1 & 2

Question 1

SpicyNo calculator
[Maximum mark: 30]

Complete the following table for the three quadratic functions below

Table to complete for three quadratic functions

Worked solution

Question 1 mark scheme

Completed table for three quadratic functions

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Question 2

MediumNo calculator
[Maximum mark: 12]

Consider the quadratic y=4x2−120x+800y=4x^2-120x+800

(a) (i) Find the roots.

(ii) Hence express the quadratic in the form y=a(x−x1)(x−x2)y=a(x-x_1)(x-x_2)[3]

(b) (i) Find the coordinates of the vertex.

(ii) Hence express the quadratic in the form y=a(x−h)2+ky=a(x-h)^2+k

(iii) Write down the equation of the axis of symmetry

(iv) Write down the minimum value of yy[5]

(c) Write down the yy-intercept of the quadratic[1]

(d) Draw the graph of the quadratic on the diagram below.[3]

Blank coordinate grid for the quadratic

Worked solution

Question 2 mark scheme

(a)

(i) x=10x=20x=10\quad x=20

(ii) y=4(x−10)(x−20)y=4(x-10)(x-20)

(b)

(i) (15,−100)(15,-100)

(ii) y=4(x−15)2−100y=4(x-15)^2-100

(iii) x=15x=15

(iv) ymin⁡=−100y_{\min}=-100

(c)

y=800y=800

(d)

Completed graph of the quadratic

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Question 3

MediumNo calculator
[Maximum mark: 14]

Consider the quadratic y=−4x2+120x−800y=-4x^2+120x-800

(a) (i) Find the roots.

(ii) Hence express the quadratic in the form y=a(x−x1)(x−x2)y=a(x-x_1)(x-x_2)[3]

(b) (i) Find the coordinates of the vertex.

(ii) Hence express the quadratic in the form y=a(x−h)2+ky=a(x-h)^2+k

(iii) Write down the equation of the axis of symmetry

(iv) Write down the maximum value of yy[5]

(c) Write down the yy-intercept of the quadratic[1]

(d) Draw the graph of the quadratic on each of the diagrams below.[5]

Two blank coordinate grids for the quadratic

Worked solution

Question 3 mark scheme

(a)

(i) x=10x=20x=10\quad x=20

(ii) y=−4(x−10)(x−20)y=-4(x-10)(x-20)

(b)

(i) (15,100)(15,100)

(ii) y=−4(x−15)2+100y=-4(x-15)^2+100

(iii) x=15x=15

(iv) ymax⁡=100y_{\max}=100

(c)

y=−800y=-800

(d)

Completed graphs of the quadratic

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Question 4

MildNo calculator
[Maximum mark: 5]

The parabola of a quadratic is shown below. The xx-intercepts are x=2x=2 and x=6x=6.

The yy-intercept is y=12y=12. The curve passes through the point A(7,5).

Parabola with x-intercepts 2 and 6 and point A at 7 comma 5

Without finding the equation of the curve and by just using the symmetry of the graph

(a) Write down the equation of the axis of symmetry.[1]

(b) Find the value of yy for x=8x=8. Justify your answer.[2]

(c) Find the value of yy for x=1x=1. Justify your answer.[2]

Worked solution

Question 4 mark scheme

(a)

x=4x=4

(b)

y=12y=12 since (8,12)(8,12) is symmetric to (0,12)(0,12) about x=4x=4

(c)

y=5y=5 since (1,5)(1,5) is symmetric to (7,5)(7,5) about x=4x=4

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Question 5

MildNo calculator
[Maximum mark: 4]

(a) Solve the equation x2−3x−10=0x^2-3x-10=0.[2]

(b) Factorize x2−3x−10x^2-3x-10.[2]

Worked solution

Question 5 mark scheme

(a)

x=5x=5 or x=−2x=-2

(b)

x2−3x−10=(x−5)(x+2)x^2-3x-10=(x-5)(x+2)

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Question 6

MildNo calculator
[Maximum mark: 4]

The diagram represents the graph of the function f:x↦(x−p)(x−q)f:x\mapsto(x-p)(x-q).

Parabola with roots negative one half and two and minimum point C

(a) Write down the values of p and q.[2]

(b) The function has a minimum value at the point C. Find the x-coordinate of C.[2]

Worked solution

Question 6 mark scheme

(a)

p=−12p=-\dfrac{1}{2}, q=2q=2 or vice versa

(b)

By symmetry C is midway between p, q ⇒\Rightarrow x-coordinate is −12+22=34\dfrac{-\frac{1}{2}+2}{2}=\dfrac{3}{4}

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Question 7

MildNo calculator
[Maximum mark: 4]

The following diagram shows part of the graph of f, where f(x)=x2−x−2f(x)=x^2-x-2.

Upward-opening parabola crossing the x-axis twice

(a) Find both x-intercepts.[2]

(b) Find the x-coordinate of the vertex.[2]

Worked solution

Question 7 mark scheme

(a)

intercepts are (−1,0)(-1,0) and (2,0)(2,0) (accept x=−1x=-1, x=2x=2)

(b)

xv=x1+x22x_v=\dfrac{x_1+x_2}{2}, xv=−b2ax_v=-\dfrac{b}{2a},

xv=0.5x_v=0.5

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Question 8

MildNo calculator
[Maximum mark: 4]

The diagram shows the parabola y=(7−x)(1+x)y=(7-x)(1+x). The points A, C are the x-intercepts and the point B is the maximum point. Find the coordinates of A, B and C.

Downward-opening parabola with x-intercepts A and C and maximum point B

Worked solution

Question 8 mark scheme

(7−x)(1+x)=0⇔x=7(7-x)(1+x)=0\Leftrightarrow x=7 or x=−1x=-1

B:x=7+−12=3;\mathrm{B}:x=\dfrac{7+-1}{2}=3;

y=(7−3)(1+3)=16y=(7-3)(1+3)=16

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Question 9

MildNo calculator
[Maximum mark: 7]

Let f(x)=8x−2x2f(x)=8x-2x^2. Part of the graph of f is shown below.

Part of a downward-opening parabola

(a) Find the x-intercepts of the graph.[4]

(b) (i) Write down the equation of the axis of symmetry.

(ii) Find the y-coordinate of the vertex.[3]

Worked solution

Question 9 mark scheme

(a)

4 and 0 (accept (4,0)(4,0) and (0,0)(0,0), or x=4x=4, x=0x=0)

(b)

(i) x=2x=2 (must be equation)

(ii) substituting x=2x=2 into f(x)f(x), y=8y=8

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Question 10

MediumNo calculator
[Maximum mark: 5]

The following diagram shows the graph of function y=ax2+bx+cy=ax^2+bx+c.

Complete the table next to the graph to show whether each expression is positive (+), negative (-) or zero (0).

Downward-opening parabola tangent to the x-axis with a table of expressions

Worked solution

Question 10 mark scheme

Completed sign table for the quadratic graph

Use the mark scheme after attempting the question yourself.

Question 11

MediumNo calculator
[Maximum mark: 5]

The following diagram shows the graph of function y=ax2+bx+cy=ax^2+bx+c.

Complete the table next to the graph to show whether each expression is positive (+), negative (-) or zero (0).

Downward-opening parabola crossing the axes with a table of expressions

Worked solution

Question 11 mark scheme

Completed sign table for the quadratic graph

Use the mark scheme after attempting the question yourself.

Question 12

MediumNo calculator
[Maximum mark: 5]

The following diagram shows the graph of function y=ax2+bx+cy=ax^2+bx+c.

Complete the table next to the graph to show whether each expression is positive (+), negative (-) or zero (0).

Upward-opening parabola crossing the axes with a table of expressions

Worked solution

Question 12 mark scheme

Completed sign table for the quadratic graph

Use the mark scheme after attempting the question yourself.

Question 13

MildNo calculator
[Maximum mark: 4]

(a) Find the vertex of f(x)=x2−6x+14f(x)=x^2-6x+14[2]

(b) Express the function in the form f(x)=(x−h)2+kf(x)=(x-h)^2+k[2]

Worked solution

Question 13 mark scheme

(a)

Vertex is (3,5)(3,5)

(b)

f(x)=(x−3)2+5f(x)=(x-3)^2+5

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Question 14

MildNo calculator
[Maximum mark: 4]

Consider the function f(x)=2x2−8x+5f(x)=2x^2-8x+5.

(a) Express f(x)f(x) in the form a(x−p)2+qa(x-p)^2+q, where a,p,q∈Za,p,q\in\mathbb{Z}.[3]

(b) Find the minimum value of f(x)f(x).[1]

Worked solution

Question 14 mark scheme

(a)

f(x)=2(x−2)2−3f(x)=2(x-2)^2-3 i.e. a=2a=2, p=2p=2, q=−3q=-3

(b)

Minimum value of f(x)=−3f(x)=-3 OR Minimum value occurs at (2,−3)(2,-3)

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Question 15

MildNo calculator
[Maximum mark: 4]

(a) Find the vertex of f(x)=2x2+2x+2f(x)=2x^2+2x+2[2]

(b) Express the function in the form f(x)=a(x−h)2+kf(x)=a(x-h)^2+k[2]

Worked solution

Question 15 mark scheme

(a)

Vertex is (−0.5,1.5)(-0.5,1.5)

(b)

f(x)=2(x+0.5)2+1.5f(x)=2(x+0.5)^2+1.5

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Question 16

MildNo calculator
[Maximum mark: 4]

(a) Find the vertex of f(x)=−x2−x−1f(x)=-x^2-x-1[2]

(b) Express the function in the form f(x)=a(x−h)2+kf(x)=a(x-h)^2+k[2]

Worked solution

Question 16 mark scheme

(a)

Vertex is (−0.5,−0.75)(-0.5,-0.75)

(b)

f(x)=−(x+0.5)2−0.75f(x)=-(x+0.5)^2-0.75

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Question 17

MediumNo calculator
[Maximum mark: 5]

Find the points of intersection between y=x2−5x+3y=x^2-5x+3 and y=3x−9y=3x-9 and sketch a graph to demonstrate the result.

Worked solution

Question 17 mark scheme

(2,−3)(2,-3) and (6,9)(6,9)

Parabola and line intersecting at two labelled points

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Question 18

MediumNo calculator
[Maximum mark: 5]

Find the point of intersection between y=x2−5x+3y=x^2-5x+3 and y=3x−13y=3x-13 and sketch a graph to demonstrate the result.

Worked solution

Question 18 mark scheme

(4,−1)(4,-1)

Parabola and tangent line meeting at one labelled point

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Question 19

MediumNo calculator
[Maximum mark: 5]

Find the points of intersection (if any) between y=x2−5x+3y=x^2-5x+3 and y=3x−15y=3x-15 and sketch a graph to demonstrate the result.

Worked solution

Question 19 mark scheme

no points of intersection

Parabola and line with no points of intersection

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Question 20

MediumNo calculator
[Maximum mark: 5]

Find the points of intersection (if any) between y=x2−3y=x^2-3 and y=5−x2y=5-x^2 and sketch a graph to demonstrate the result.

Worked solution

Question 20 mark scheme

(−2,1)(-2,1) and (2,1)(2,1)

Two parabolas intersecting at two labelled points

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