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Quadratics

Practice exam-style IB Math AA questions for Quadratics, aligned with the syllabus and grouped by topic.

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Question 1
SL • Paper 1
Easy
Non Calculator
SL • Paper 1
Easy
Non Calculator

A quadratic function has xx-intercepts (−2,0)(-2,0) and (4,0)(4,0). Its graph passes through the point (1,−9)(1,-9).

A

the function in factorized form.

[3]
B

the equation of the axis of symmetry.

[1]
C

the yy-intercept of the graph of ff.

[1]
Question 2
SL • Paper 2
Easy
Calculator Permitted
SL • Paper 2
Easy
Calculator Permitted

The quadratic function ff is given by

f(x)=2x2−12x+11f(x)=2x^2-12x+11
A

Write f(x)f(x) in the form a(x−h)2+ka(x-h)^2+k.

[2]
B

the coordinates of the vertex of the graph of y=f(x)y=f(x).

[1]
C

the two zeros of ff.

[2]
Question 3
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

Consider the quadratic function f(x)=2x2−12x+13f(x)=2x^2-12x+13, for x∈Rx\in\mathbb{R}.

A

Express f(x)f(x) in the form a(x−h)2+ka(x-h)^2+k.

[2]
B

the coordinates of the vertex of the graph of ff.

[1]
C

the inequality f(x)≤3f(x)\leq 3.

[3]
Question 4
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

The height, hh metres, of a ball above the ground tt seconds after it is thrown is modelled by
h(t)=−5t2+20t+25h(t)=-5t^2+20t+25, where t≥0t\geq 0.

A

Express h(t)h(t) in vertex form.

[2]
B

the maximum height reached by the ball and the time at which it occurs.

[2]
C

the time at which the ball first reaches the ground.

[1]

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Question 5
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

Consider the quadratic equation x2−2kx+3k−2=0x^2-2kx+3k-2=0, where k∈Rk\in\mathbb{R}.

A

the discriminant in terms of kk, giving your answer in factorized form.

[2]
B

the values of kk for which the equation has two equal real roots.

[1]
C

the values of kk for which the equation has no real roots.

[2]
Question 6
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

Consider the inequality (x−1)(x−5)≤3(x-1)(x-5)\leq 3, where x∈Rx\in\mathbb{R}.

A

the roots of the equation (x−1)(x−5)=3(x-1)(x-5)=3.

[2]
B

the inequality.

[2]
Question 7
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

A ball is thrown vertically upwards from the top of a platform. Its height above the ground, hh metres, after tt seconds is modelled by

h(t)=−4.9t2+18t+12,t≥0h(t)=-4.9t^2+18t+12, \quad t\ge0
A

the greatest height reached by the ball.

[2]
B

the time at which the ball reaches the ground.

[2]
C

the time interval during which the ball is at least 2020 metres above the ground.

[2]
Question 8
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

Consider the quadratic equation

(k+2)x2−4kx+(k−1)=0(k+2)x^2-4kx+(k-1)=0

where k∈Rk\in\mathbb{R} and k≠−2k\ne -2.

A

an expression for the discriminant in terms of kk.

[2]
B

the nature of the roots of the equation for all possible values of kk.

[3]

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Question 9
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

The graph of y=x2−5x−6y=x^2-5x-6 is shown.

A

the xx-intercepts of the graph.

[2]
B

the inequality x2−5x−6<0x^2-5x-6<0.

[2]
C

the minimum value of x2−5x−6x^2-5x-6.

[1]
Question 10
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

The quadratic equation x2−mx+m+3=0x^2-mx+m+3=0 has roots α\alpha and β\beta. One root is twice the other.

A

α+β\alpha+\beta and αβ\alpha\beta in terms of mm.

[2]
B

the possible values of mm.

[4]
Question 11
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

For k∈Rk\in\mathbb{R}, consider the function fk(x)=x2+2kx+k+3f_k(x)=x^2+2kx+k+3, where x∈Rx\in\mathbb{R}.

A

the minimum value of fk(x)f_k(x) in terms of kk.

[2]
B

the values of kk for which fk(x)>0f_k(x)>0 for all real xx.

[3]
Question 12
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

For a∈Ra\in\mathbb{R}, let fa(x)=x2−2(a+1)x+a2+4af_a(x)=x^2-2(a+1)x+a^2+4a.

A

Express fa(x)f_a(x) in vertex form.

[2]
B

The vertex of the graph of faf_a is (H,K)(H,K). a linear relation between HH and KK.

[2]
C

the values of aa for which the graph of faf_a has two distinct xx-intercepts.

[2]

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Question 13
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

A quadratic function ff has a vertex at (2,−3)(2,-3). Its two xx-intercepts are 66 units apart, and the coefficient of x2x^2 is positive.

A

the two xx-intercepts of the graph of ff.

[2]
B

f(x)f(x) in the form a(x−p)(x−q)a(x-p)(x-q).

[3]
Question 14
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

The line y=mx+1y=mx+1 intersects the parabola y=x2−2x+5y=x^2-2x+5, where m∈Rm\in\mathbb{R}.

A

the quadratic equation in xx whose roots give the xx-coordinates of the intersection points.

[1]
B

the values of mm for which the line is tangent to the parabola.

[2]
C

the values of mm for which the line does not intersect the parabola.

[2]
Question 15
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

A rectangular garden is to be built against a straight wall. Fencing is needed for the other three sides only. The total length of fencing available is 3030 metres. The side parallel to the wall has length xx metres and each perpendicular side has length yy metres.

A

an equation connecting xx and yy.

[1]
B

the area A m2A\ \text{m}^2 of the garden is given by A=15x−12x2A=15x-\frac{1}{2}x^2.

[2]
C

the maximum possible area of the garden.

[3]
Question 16
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

A quadratic model y=ax2+bx+cy=ax^2+bx+c is fitted to three measured points, shown in the table.

x

y

1

5.5

3

2.5

6

11.5

A

Using the data in the table, the quadratic model in the form y=ax2+bx+cy=ax^2+bx+c.

[3]
B

the minimum value predicted by this model.

[2]
C

the value of xx at which this minimum occurs.

[1]

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Question 17
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

The line y=mx+1y=mx+1 is tangent to the parabola y=x2−4x+7y=x^2-4x+7.

A

mm satisfies m2+8m−8=0m^2+8m-8=0.

[3]
B

the possible values of mm.

[2]
C

For the positive value of mm, the point of tangency.

[1]
Question 18
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

The quadratic equation

x2−6x+k=0x^2-6x+k=0

has two distinct real roots α\alpha and β\beta.

A

α+β\alpha+\beta and αβ\alpha\beta in terms of kk.

[1]
B

Given that α2+β2=20\alpha^2+\beta^2=20, the value of kk.

[2]
C

For this value of kk, form the quadratic equation with roots 1α\frac{1}{\alpha} and 1β\frac{1}{\beta}.

[2]
Question 19
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

The graph of a quadratic function ff has vertex (2,−5)(2,-5) and passes through the point (7,20)(7,20).

A

f(x)f(x) in the form a(x−h)2+ka(x-h)^2+k.

[3]
B

the exact xx-intercepts of the graph of ff.

[2]
C

the distance between the two xx-intercepts.

[1]
Question 20
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

The quadratic function ff has vertex (2,−9)(2,-9) and passes through the point (0,3)(0,3).

A
I.

f(x)f(x) in the form a(x−h)2+ka(x-h)^2+k.

[2]
II.

express f(x)f(x) in the form ax2+bx+cax^2+bx+c.

[2]
B
I.

the xx-intercepts of the graph of y=f(x)y=f(x). If you did not obtain f(x)=3(x−2)2−9f(x)=3(x-2)^2-9, use this expression.

[2]
II.

the inequality f(x)<3f(x)<3. If you did not obtain f(x)=3(x−2)2−9f(x)=3(x-2)^2-9, use this expression.

[2]
C

the range of ff for the restricted domain 0≤x≤50\le x\le 5.

[2]

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Question 21
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

The height, hh metres, of a small rocket above the ground tt seconds after launch is modelled by

h(t)=−4t2+16t+5,t≥0h(t)=-4t^2+16t+5,\quad t\ge 0
A
I.

Express h(t)h(t) in the form a(t−p)2+qa(t-p)^2+q.

[2]
II.

the maximum height of the rocket and the time at which it occurs.

[2]
B

the time at which the rocket reaches the ground. If you did not obtain h(t)=−4(t−2)2+21h(t)=-4(t-2)^2+21, use this expression.

[3]
C
I.

the interval of time for which the rocket is at least 1717 metres above the ground. If you did not obtain h(t)=−4(t−2)2+21h(t)=-4(t-2)^2+21, use this expression.

[2]
II.

the second time at which the rocket is at its launch height.

[1]
Question 22
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

An arch is modelled by a parabola. The height of the arch, HH metres, above a horizontal floor is given by a quadratic function of the horizontal distance xx metres from the left end. The arch meets the floor at x=0x=0 and x=12x=12, and its greatest height is 1818 metres.

A
I.

the equation of the axis of symmetry of the arch.

[1]
II.

the height of the arch may be written as H=12x(12−x)H=\frac12 x(12-x).

[3]
B

A horizontal beam is placed at a height of 1010 metres above the floor. the horizontal width of the arch at this height. If you did not obtain H=12x(12−x)H=\frac12 x(12-x), use this expression.

[4]
C

A rectangular vehicle is 66 metres wide and passes centrally under the arch. the greatest possible height of the vehicle.

[2]
Question 23
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

A quadratic function ff has xx-intercepts (−1,0)(-1,0) and (5,0)(5,0), and passes through the point (1,−12)(1,-12).

A
I.

f(x)f(x) in factorized form.

[2]
II.

the coordinates of the vertex of the graph of y=f(x)y=f(x).

[3]
B

the inequality f(x)≤0f(x)\le 0. If you did not obtain f(x)=32(x+1)(x−5)f(x)=\frac32(x+1)(x-5), use this expression.

[2]
C

The line through the vertex and the yy-intercept of the graph of ff is denoted by LL. the equation of LL.

[3]
Question 24
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

A quadratic model y=ax2+bx+cy=ax^2+bx+c passes through the three points (0,5)(0,5), (2,−3)(2,-3) and (5,0)(5,0).

1

2

3

x

0

2

5

y

5

-3

0

A
I.

the value of cc.

[1]
II.

the values of aa and bb.

[3]
B

Express the model in vertex form and its minimum value. If you did not obtain y=x2−6x+5y=x^2-6x+5, use this expression.

[3]
C

the values of xx for which the model predicts y≤0y\le 0.

[2]

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Question 25
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

The quadratic equation (k−1)x2−2kx+k+3=0(k-1)x^2-2kx+k+3=0, where k≠1k\neq 1, has roots α\alpha and β\beta. The roots differ by 22.

A

α+β\alpha+\beta and αβ\alpha\beta in terms of kk.

[2]
B

the possible values of kk.

[4]
Question 26
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Consider the quadratic equation x2−2(k+1)x+k2−1=0x^2-2(k+1)x+k^2-1=0, where k∈Rk\in\mathbb{R}.

A

the discriminant in terms of kk.

[2]
B

the values of kk for which the equation has two distinct positive real roots.

[4]
Question 27
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

The cross-section of a pedestrian arch is modelled by a parabola. The arch meets the ground at points A(0,0)A(0,0) and B(12,0)B(12,0), where distances are measured in metres. The greatest height of the arch is 55 metres.

A
I.

the equation of the axis of symmetry of the parabola.

[1]
II.

an equation for the height hh of the arch in terms of the horizontal distance xx from AA.

[4]
B
I.

the two values of xx for which the height of the arch is 3.23.2 metres.

[2]
II.

the horizontal width of the part of the arch that is at least 3.23.2 metres high.

[2]
C

A rectangular vehicle of height HH metres is driven centrally through the arch. the maximum possible width ww metres of the vehicle is given by w=121−H5w=12\sqrt{1-\dfrac{H}{5}}, for 0≤H≤50\leq H\leq5.

[2]
Question 28
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

A company sells headphones. If xx headphones are produced and sold in one week, the selling price is modelled by p=50−0.025xp=50-0.025x dollars per headphone. The weekly cost, in dollars, is modelled by C=900+18xC=900+18x. The factory can produce at most 10001000 headphones per week.

A
I.

an expression for the weekly revenue RR in terms of xx.

[1]
II.

the weekly profit PP is given by P=−0.025x2+32x−900P=-0.025x^2+32x-900.

[2]
B
I.

the number of headphones that should be produced and sold to maximize the weekly profit.

[2]
II.

the maximum weekly profit.

[2]
C
I.

the break-even values of xx, according to the model.

[2]
II.

Taking account of the factory capacity and the fact that xx must be a whole number, the values of xx for which the company makes a profit.

[2]

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Question 29
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

A sensor records a signal strength SS at time tt seconds. The signal is modelled by a quadratic function S(t)=at2+bt+cS(t)=at^2+bt+c, for 0≤t≤90\leq t\leq9. Three recorded values are shown in the table.

time t [s]

signal strength S

0

2.1

4

5.7

9

1.2

A
I.

Use the data to the values of aa, bb and cc.

[3]
II.

Write the model in vertex form.

[1]
B
I.

the maximum signal strength predicted by the model.

[1]
II.

the times at which the signal strength is 44.

[2]
C

A technician wants the signal strength to be at least a threshold value TT for exactly 44 seconds. the greatest possible value of TT.

[4]
Question 30
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

A farmer uses 8080 metres of fencing to make three identical rectangular pens side by side. Each pen has length xx metres and width yy metres. The pens share two internal fences, each of length yy metres.

A
I.

6x+4y=806x+4y=80.

[1]
II.

the total area AA of the three pens is given by A=60x−92x2A=60x-\dfrac{9}{2}x^2.

[3]
B
I.

the value of xx that gives the maximum total area.

[1]
II.

the corresponding value of yy and the maximum total area.

[2]
C

The farmer changes the design to make nn identical pens side by side, using the same 8080 metres of fencing. the maximum possible total area in terms of nn.

[3]
Question 31
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

Two functions are defined by f(x)=x2−6x+cf(x)=x^2-6x+c and g(x)=−12x2+4x+1g(x)=-\dfrac{1}{2}x^2+4x+1, where cc is a real constant.

A
I.

the xx-coordinates of any points of intersection satisfy 1.5x2−10x+c−1=01.5x^2-10x+c-1=0.

[2]
II.

the discriminant of this quadratic equation in terms of cc.

[1]
B
I.

the value of cc for which the graphs are tangent.

[2]
II.

the number of points of intersection when c=20c=20.

[1]
C

For c=5c=5, the inequality f(x)≤g(x)f(x)\leq g(x), giving your answer in interval notation.

[4]
Question 32
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

A decorative panel is made in the shape of a parabola with vertical axis of symmetry. In a coordinate system with origin at the lowest point of the panel, the curve passes through the points (−2.4,1.8)(-2.4,1.8) and (2.4,1.8)(2.4,1.8), where distances are in metres.

A
I.

why the equation of the curve may be written in the form y=ax2y=ax^2.

[1]
II.

the value of aa.

[2]
III.

the equation of the curve.

[1]
B
I.

the width of the panel at a height of 1.01.0 metre above the lowest point.

[2]
II.

the height of the panel at a horizontal distance of 1.51.5 metres from the axis of symmetry.

[1]
C

A rectangular sign of height 0.800.80 metres is placed centrally with its lower edge at the lowest point of the panel. the greatest possible width of the sign.

[3]

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Question 33
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Consider the equation

(k−1)x2+2kx+k+3=0(k-1)x^2+2kx+k+3=0

where k∈Rk\in\mathbb{R}.

A

the values of kk for which the equation is quadratic and has two equal real roots.

[4]
B

For k=1k=1, the resulting equation.

[2]
Question 34
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

The graphs of

y=x2−3x+1y=x^2-3x+1

and

y=2x+ky=2x+k

intersect at two points whose xx-coordinates differ by 44.

A

the xx-coordinates of the points of intersection satisfy x2−5x+(1−k)=0x^2-5x+(1-k)=0.

[2]
B

the value of kk.

[3]
C

For this value of kk, the coordinates of the two points of intersection.

[2]
Question 35
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

A family of quadratic functions is defined by

fk(x)=x2−2kx+3kf_k(x)=x^2-2kx+3k

where k∈Rk\in\mathbb{R}.

A

the coordinates of the vertex of the graph of y=fk(x)y=f_k(x) in terms of kk.

[2]
B

the values of kk for which the graph of y=fk(x)y=f_k(x) does not meet the xx-axis.

[3]
C

For the values of kk found in part (b), the greatest possible value of the yy-coordinate of the vertex.

[2]
Question 36
SL • Paper 1
Hard
Non Calculator
SL • Paper 1
Hard
Non Calculator

Consider the quadratic equation

x2−2(k−1)x+k2−5=0x^2-2(k-1)x+k^2-5=0

where k∈Rk\in\mathbb{R}.

A
I.

the discriminant in terms of kk, giving your answer in factorized form.

[3]
II.

the values of kk for which the equation has two distinct real roots, two equal real roots, and no real roots.

[2]
B

the equation when k=1k=1.

[2]
C

the values of kk for which both roots of the equation are real and positive.

[4]

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Question 37
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

For t∈Rt\in\mathbb{R}, consider the family of quadratic functions

ft(x)=x2−2tx+4t−3f_t(x)=x^2-2tx+4t-3
A
I.

the coordinates of the vertex of the graph of y=ft(x)y=f_t(x) in terms of tt.

[3]
II.

The vertex is denoted by (H,K)(H,K). K=−(H−2)2+1K=-(H-2)^2+1.

[2]
B

the values of tt for which the graph of y=ft(x)y=f_t(x) has two distinct xx-intercepts.

[3]
C

the values of tt for which ft(x)>0f_t(x)>0 for all x∈Rx\in\mathbb{R}.

[2]
Question 38
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

The line LkL_k has equation y=kx+1y=kx+1, where k∈Rk\in\mathbb{R}. The parabola PP has equation y=x2−4x+7y=x^2-4x+7.

A
I.

the xx-coordinates of the points of intersection of LkL_k and PP satisfy x2−(k+4)x+6=0x^2-(k+4)x+6=0.

[2]
II.

the values of kk for which LkL_k is tangent to PP.

[3]
B

For k=−4+26k=-4+2\sqrt{6}, the point of tangency. If you did not obtain x2−(k+4)x+6=0x^2-(k+4)x+6=0, use this equation.

[3]
C

the values of kk for which LkL_k intersects PP at two points whose xx-coordinates are both positive.

[3]
Question 39
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

The quadratic equation

x2−kx+(k+1)=0x^2-kx+(k+1)=0

has roots α\alpha and β\beta, where k∈Rk\in\mathbb{R}.

A
I.

α+β\alpha+\beta and αβ\alpha\beta in terms of kk.

[2]
II.

Given that α2+β2=6\alpha^2+\beta^2=6, the possible values of kk.

[4]
B

only one of the values of kk found in part (a)(ii) gives real roots for the original equation.

[2]
C

For the value of kk which gives real roots, form the quadratic equation with roots 1α\frac{1}{\alpha} and 1β\frac{1}{\beta}.

[3]
Question 40
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

For k∈Rk\in\mathbb{R}, consider the family of quadratic functions fk(x)=x2−2kx+k2−4k+1f_k(x)=x^2-2kx+k^2-4k+1. The graph of y=fk(x)y=f_k(x) has vertex VkV_k.

A
I.

Express fk(x)f_k(x) in vertex form.

[2]
II.

the coordinates of VkV_k and the equation of the locus of VkV_k.

[2]
B
I.

the values of kk for which the graph has two distinct xx-intercepts.

[2]
II.

For k>14k>\dfrac{1}{4}, the distance between the two xx-intercepts is 24k−12\sqrt{4k-1}.

[2]
C

The two xx-intercepts and VkV_k form a triangle. the value of kk for which the area of this triangle is 2727.

[4]

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Question 41
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The quadratic equation x2−(2k+1)x+k2−3=0x^2-(2k+1)x+k^2-3=0 has roots α\alpha and β\beta, where k∈Rk\in\mathbb{R}.

A
I.

α+β\alpha+\beta and αβ\alpha\beta in terms of kk.

[2]
II.

the discriminant in terms of kk.

[2]
B
I.

the values of kk for which the equation has two distinct positive real roots.

[3]
C

Given that α2+β2=30\alpha^2+\beta^2=30, the value of kk for which the equation has real roots.

[4]
Question 42
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The graphs of y=x2+ax+4y=x^2+ax+4 and y=−x2+2x+ay=-x^2+2x+a intersect at points whose xx-coordinates satisfy a quadratic equation. Here a∈Ra\in\mathbb{R}.

A
I.

the xx-coordinates of the points of intersection satisfy 2x2+(a−2)x+(4−a)=02x^2+(a-2)x+(4-a)=0.

[2]
II.

the discriminant of this equation in terms of aa.

[1]
B
I.

the values of aa for which the graphs are tangent.

[2]
II.

the values of aa for which the graphs intersect at two distinct points.

[2]
C

For a=6a=6, the coordinates of the two points of intersection and the distance between them.

[5]
Question 43
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A family of quadratic functions is defined by

ft(x)=x2−2tx+t2−4t+3f_t(x)=x^2-2tx+t^2-4t+3

where t∈Rt\in\mathbb{R}. The graph of y=ft(x)y=f_t(x) has vertex VtV_t. The diagram shows several members of the family and the line on which their vertices appear to lie.

A
I.

Express ft(x)f_t(x) in vertex form and the coordinates of VtV_t.

[2]
II.

the vertices of all graphs in the family lie on a straight line.

[1]
B

the values of tt for which the graph of y=ft(x)y=f_t(x) has two distinct xx-intercepts.

[3]
C
I.

the line found in part (a)(ii) intersects the graph of y=ft(x)y=f_t(x) at x=tx=t and x=t−4x=t-4.

[2]
II.

the distance between the two intersection points, showing that it is independent of tt. If you did not obtain the two values in part (c)(i), use x=tx=t and x=t−4x=t-4.

[1]
Question 44
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A stream of water from a fountain is modelled by

hk(x)=−0.2x2+kx+1h_k(x)=-0.2x^2+kx+1

where xx is the horizontal distance in metres from the nozzle, hk(x)h_k(x) is the height in metres above the ground, and k>0k>0 is a parameter controlled by the nozzle angle. A low wall is located 44 metres from the nozzle and has height 33 metres.

A

Express hk(x)h_k(x) in vertex form and the maximum height in terms of kk.

[3]
B
I.

the least value of kk for which the stream clears the wall.

[2]
II.

For k=1.3k=1.3, the horizontal distance from the nozzle to the splash point.

[2]
C

The fountain is redesigned so that the stream must clear the wall and land between 88 m and 1010 m from the nozzle, inclusive. the possible values of kk.

[4]

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Question 45
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A simplified model for the temperature TT degrees Celsius in a laboratory between 02:00 and 08:00 is

Tm(x)=m(x−2)(x−8)+20T_m(x)=m(x-2)(x-8)+20

where xx is the time in hours after midnight and m>0m>0. The graph is used to the time interval during which the temperature is at or below a specified level. The graph shows the curves from x=0x=0 to x=10x=10; portions with x<2x<2 or x>8x>8 are extrapolations outside the model's stated validity interval.

A
I.

Expand Tm(x)T_m(x) into standard form.

[1]
II.

the equation of the axis of symmetry and the minimum temperature in terms of mm.

[2]
B

For m=1.5m=1.5, the times at which the temperature is 12∘C12^\circ\text{C}, giving your answers to the nearest minute.

[3]
C

the value of mm for which the temperature is at or below 12∘C12^\circ\text{C} for exactly 22 hours.

[3]
D

For a general threshold temperature HH, where H<20H<20, the duration DD hours for which Tm(x)≤HT_m(x)\le H is

D=29−20−HmD=2\sqrt{9-\frac{20-H}{m}}

and the condition on mm for this duration to exist.

[3]
Question 46
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The fixed parabola

y=x2−4x+1y=x^2-4x+1

is intersected by the family of lines

y=mx−3y=mx-3

where m∈Rm\in\mathbb{R}. The diagram shows some possible intersections.

A
I.

the xx-coordinates of the intersection points satisfy

x2−(m+4)x+4=0x^2-(m+4)x+4=0
[2]
II.

the values of mm for which the line intersects the parabola at two distinct points.

[2]
B

For m=5m=5, the exact xx-coordinates of the intersection points.

[2]
C
I.

the square of the horizontal distance between the two intersection points is (m+4)2−16(m+4)^2-16.

[2]
II.

the values of mm for which the horizontal distance between the intersection points is 33. If you did not obtain part (c)(i), use (m+4)2−16(m+4)^2-16 for the square of the distance.

[2]
D

the two tangent lines from the family and their points of tangency.

[2]
Question 47
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A rectangular courtyard is 3030 m by 1818 m. A uniform paved border of width xx metres is built around the inside edge, leaving a central rectangular grass region of area A(x)A(x) square metres. The diagram shows the courtyard and grass region.

A
I.

A(x)=4x2−96x+540A(x)=4x^2-96x+540.

[2]
II.

the possible values of xx.

[1]
III.

the axis of symmetry of the graph of A(x)A(x).

[1]
B

the width of the border when the grass area is 320 m2320\ \text{m}^2.

[3]
C

the values of xx for which the grass area is at least 320 m2320\ \text{m}^2.

[2]
D
I.

For a general rectangular courtyard of dimensions LL by WW, where L>W>0L>W>0, the grass area is a decreasing function of xx throughout its physical domain.

[3]
II.

, in terms of LL, WW and BB, the greatest possible border width if the grass area must be at least BB, where 0<B<LW0<B<LW.

[2]
Question 48
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

The parabolas C1C_1 and C2C_2 have equations

C1:y=x2−6x+5,C2:y=−x2+2kx−1C_1: y=x^2-6x+5,\qquad C_2: y=-x^2+2kx-1

where k∈Rk\in\mathbb{R}.

A
I.

the xx-coordinates of the points of intersection satisfy x2−(k+3)x+3=0x^2-(k+3)x+3=0.

[2]
II.

the values of kk for which the two parabolas are tangent to each other.

[3]
B

For k=−3+23k=-3+2\sqrt{3}, the point of tangency. If you did not obtain x2−(k+3)x+3=0x^2-(k+3)x+3=0, use this equation.

[3]
C

the values of kk for which the two parabolas intersect at two points whose xx-coordinates are both greater than 11.

[4]

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Question 49
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

For a∈Ra\in\mathbb{R}, a≠0a\ne 0, consider the quadratic function

fa(x)=ax2−2(a+1)x+a+4f_a(x)=ax^2-2(a+1)x+a+4
A
I.

the xx-coordinate of the vertex in terms of aa.

[2]
II.

The vertex is denoted by (H,K)(H,K). H+K=3H+K=3.

[2]
B

the values of aa for which the graph of y=fa(x)y=f_a(x) opens upwards and its vertex is below the xx-axis.

[3]
C

the values of aa for which the graph of y=fa(x)y=f_a(x) has no xx-intercepts.

[2]
D

For a=14a=\frac14, fa(x)≤0f_a(x)\le 0.

[3]
Question 50
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

For p∈Rp\in\mathbb{R}, consider the quadratic function

qp(x)=x2−2px+p2−pq_p(x)=x^2-2px+p^2-p
A
I.

the coordinates of the vertex of the graph of y=qp(x)y=q_p(x).

[2]
II.

the roots of qp(x)=0q_p(x)=0 in terms of pp, stating the condition for the roots to be real.

[3]
B

the values of pp for which the equation qp(x)=0q_p(x)=0 has two distinct positive roots. If you did not obtain the roots p±pp\pm\sqrt{p}, use these roots.

[3]
C

Let the two roots be α\alpha and β\beta. the value of pp for which 1α+1β=4\frac{1}{\alpha}+\frac{1}{\beta}=4, given that the roots are positive.

[3]
D

the values of pp for which the number 11 lies strictly between the two roots of qp(x)=0q_p(x)=0.

[2]
Question 51
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

For m∈Rm\in\mathbb{R}, define qm(x)=x2−2mx+m+2q_m(x)=x^2-2mx+m+2. A designer wants qm(x)q_m(x) to be non-negative for every real value of xx in a model.

A
AI.

Express qm(x)q_m(x) in vertex form.

[2]
AII.

the minimum value of qm(x)q_m(x) in terms of mm.

[2]
B
BI.

the values of mm for which qm(x)≥0q_m(x)\geq0 for all x∈Rx\in\mathbb{R}.

[3]
BII.

For which values of mm does qm(x)=0q_m(x)=0 have exactly one real solution?

[1]
C

The designer instead only requires qm(x)≥0q_m(x)\geq0 for all xx in the interval 0≤x≤40\leq x\leq4. the set of possible values of mm.

[4]
Question 52
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A family of parabolas is given by ft(x)=t(x−1)(x−5)+8f_t(x)=t(x-1)(x-5)+8, where tt is a non-zero real parameter. The vertex of the graph of y=ft(x)y=f_t(x) is denoted by VtV_t.

A
I.

the equation of the axis of symmetry of the graph of y=ft(x)y=f_t(x).

[1]
II.

the coordinates of VtV_t in terms of tt.

[3]
B
I.

the values of tt for which the graph has two distinct xx-intercepts.

[3]
II.

the value of tt for which the graph touches the xx-axis.

[1]
C

For t<0t<0 or t>2t>2, let the two xx-intercepts be PP and QQ. the length PQPQ is 24−8t2\sqrt{4-\dfrac{8}{t}}.

[4]

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Question 53
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

For r∈Rr\in\mathbb{R}, consider the quadratic equation (r+1)x2−2rx+r−3=0(r+1)x^2-2rx+r-3=0. When r=−1r=-1, the equation is not quadratic.

A
I.

For r≠−1r\neq-1, the discriminant in terms of rr.

[2]
II.

the values of rr for which the equation is quadratic and has two distinct real roots.

[2]
B
I.

For r≠−1r\neq-1, the sum and product of the roots in terms of rr.

[2]
II.

the values of rr for which the equation has two distinct positive real roots.

[3]
C

For the case r=−1r=-1, the resulting equation. why it must be considered separately from the discriminant analysis.

[4]
Question 54
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A quadratic curve y=fa(x)y=f_a(x) passes through the two fixed points P(−1,6)P(-1,6) and Q(3.5,1.5)Q(3.5,1.5). It is written in the form

fa(x)=ax2+bx+cf_a(x)=ax^2+bx+c

where a≠0a\ne0. The diagram shows several possible curves through PP and QQ.

A
I.

b=−1−52ab=-1-\frac52a and c=5−72ac=5-\frac72a.

[3]
II.

Write fa(x)f_a(x) in terms of aa only.

[1]
B

the values of aa for which the graph is tangent to the xx-axis.

[4]
C
I.

For each value of aa found in part (b), the point of tangency with the xx-axis. If you did not obtain values in part (b), use a=227a=\frac{2}{27} and a=23a=\frac23.

[2]
II.

the values of aa for which the graph has no real xx-intercepts.

[2]
Question 55
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A vertical motion model is given by

sv(t)=−t2+vt+4−vs_v(t)=-t^2+vt+4-v

where sv(t)s_v(t) is the height in metres after tt seconds, t≥0t\ge0, and vv is a real parameter. A horizontal beam is located at height 99 metres.

A
I.

Write sv(t)s_v(t) in vertex form.

[2]
II.

the maximum height in terms of vv, assuming the vertex occurs for t≥0t\ge0.

[2]
B

the values of vv for which the object reaches the beam at least once.

[4]
C
I.

For v=8v=8, the two times at which the object is at height 99 metres.

[2]
II.

the value of vv for which the object just touches the beam, and the time at which this occurs. If you did not obtain part (b), use v2−4v−20=0v^2-4v-20=0.

[3]
Question 56
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For 0<d<50<d<5, a quadratic function has xx-intercepts at 5−d5-d and 5+d5+d, and has yy-intercept 2020. It is denoted by fdf_d. The diagram shows several such parabolas for different values of dd.

A
I.

fd(x)=2025−d2(x−(5−d))(x−(5+d))f_d(x)=\frac{20}{25-d^2}(x-(5-d))(x-(5+d))
[2]
II.

whether the parabola opens upwards or downwards.

[1]
III.

the equation of the axis of symmetry.

[1]
B

the minimum value of fd(x)f_d(x) in terms of dd.

[2]
C
I.

dd if the minimum value is −45-45.

[2]
II.

For this value of dd, write fd(x)f_d(x) in vertex form.

[2]
D

the possible values of dd for which the minimum value is less than −80-80.

[3]

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Question 57
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For each real number pp, consider the quadratic equation

x2−2px+p+2=0x^2-2px+p+2=0

When the roots are real, they are denoted by α\alpha and β\beta. The graph shows how the number of real roots changes as pp varies.

A
I.

the discriminant in terms of pp.

[1]
II.

the values of pp for which the equation has two distinct real roots.

[2]
III.

the values of pp for which the equation has equal real roots.

[1]
B

, when the roots are real,

(α−β)2=4(p2−p−2)(\alpha-\beta)^2=4(p^2-p-2)
[3]
C
I.

the values of pp for which the roots differ by 66. If you did not obtain part (b), use (α−β)2=4(p2−p−2)(\alpha-\beta)^2=4(p^2-p-2).

[3]
II.

that there is no value of pp for which the roots are real and both roots lie strictly between 00 and 11.

[2]
Question 58
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A family of parabolas is given by

y=a(x−1)2+2y=a(x-1)^2+2

where a>0a>0. A point MM lies on the graph, and its coordinates are (x,y)(x,y). The area of the rectangle with opposite vertices (1,2)(1,2) and MM is studied for points on the right-hand branch of the parabola.

A
I.

Let u=x−1u=x-1, where u>0u>0. Express the vertical side length of the rectangle in terms of uu and aa.

[1]
II.

write the area RR of the rectangle in terms of uu and aa.

[1]
III.

For a=12a=\frac12, xx when R=32R=32.

[2]
B

A horizontal line y=18y=18 intersects the parabola in two points. the value of aa for which the distance between these two points is 88.

[3]
C
I.

the line through (1,2)(1,2) and MM has gradient auau.

[2]
II.

For a=2a=2, the point MM for which this gradient is 1010. If you did not obtain part (c)(i), use gradient auau.

[2]
D

a relationship between the area RR of the rectangle and the gradient gg of the line through (1,2)(1,2) and MM, in terms of aa.

[3]
Question 59
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The concentration CC of a medicine in a patient, in arbitrary units, is modelled for 0≤t≤80\le t\le 8 by

Ck(t)=−kt2+6kt+2C_k(t)=-kt^2+6kt+2

where k>0k>0 and tt is measured in hours. The medicine is considered effective when Ck(t)≥10C_k(t)\ge 10.

A
I.

Express Ck(t)C_k(t) in vertex form.

[2]
II.

the maximum concentration in terms of kk.

[1]
III.

the least value of kk for which the medicine is ever effective.

[1]
B

For k=2k=2, the time interval during which the medicine is effective.

[3]
C
I.

, for k>89k>\frac89, the duration DD for which the medicine is effective is

D=29−8kD=2\sqrt{9-\frac8k}
[2]
II.

the value of kk for which the medicine is effective for exactly 44 hours. If you did not obtain part (c)(i), use D=29−8kD=2\sqrt{9-\frac8k}.

[2]
D

Evaluate the model by determining whether it is possible for the medicine to be effective for the entire interval 0≤t≤80\le t\le8.

[2]
Question 60
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

Two families of parabolas are defined by

fa(x)=x2+ax+1,ga(x)=−x2+2x+af_a(x)=x^2+ax+1,\qquad g_a(x)=-x^2+2x+a

where a∈Ra\in\mathbb{R}. The intersections of their graphs depend on aa.

A
I.

the xx-coordinates of the intersection points satisfy

2x2+(a−2)x+(1−a)=02x^2+(a-2)x+(1-a)=0
[2]
II.

the discriminant of this quadratic in terms of aa.

[2]
III.

the values of aa for which the two graphs intersect at two distinct points.

[1]
B

For the value of aa for which the graphs are tangent and a>−2a>-2, the point of tangency.

[4]
C
I.

Let the two intersection xx-coordinates be rr and ss.

r+s=2−a2,rs=1−a2r+s=\frac{2-a}{2},\qquad rs=\frac{1-a}{2}
[2]
II.

the values of aa for which the two intersection xx-coordinates have product −3-3. If you did not obtain part (c)(i), use rs=1−a2rs=\frac{1-a}{2}.

[2]
D

the values of aa for which both intersection points are distinct and have positive xx-coordinates.

[2]

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Properties of Functions

Rational Functions