# IBDP Maths AI: Topic: SL 2.3: The graph of a function: IB style Questions SL Paper 1

## Question

The function $$f(x) = 5 – 3({2^{ – x}})$$ is defined for $$x \geqslant 0$$.

On the axes below sketch the graph of f (x) and show the behaviour of the curve as x increases.[3]

a.i.

Write down the coordinates of any intercepts with the axes.

[1]

a.ii.

Draw the line y = 5 on your sketch.[1]

b.

Write down the number of solutions to the equation f (x) = 5 .[1]

c.

## Markscheme

(A1)(A1)(A1)

Notes: Award (A1) for labels and scale on y-axis.
Award (A1) for smooth increasing curve in the given domain.
Award (A1) for asymptote implied ($${\text{gradient}} \to {\text{0}}$$).[3 marks]

a.i.

(0, 2) accept x = 0, y = 2     (A1)     (C4)

Note: If incorrect domain used and both (0, 2) and ($$–$$0.737, 0) seen award (A1)(ft).[1 mark]

a.ii.

line passing through (0, 5), parallel to x axis and not intersecting their graph.     (A1)     (C1)[1 mark]

b.

zero     (A1)     (C1)[1 mark]

c.

## Question

The $$x$$-coordinate of the minimum point of the quadratic function $$f(x) = 2{x^2} + kx + 4$$ is $$x =1.25$$.

(i) Find the value of $$k$$ .

(ii) Calculate the $$y$$-coordinate of this minimum point.[4]

a.

Sketch the graph of $$y = f(x)$$ for the domain $$– 1 \leqslant x \leqslant 3$$.

[2]

b.

## Markscheme

(i)     $$1.25 = – \frac{k}{{2(2)}}$$     (M1)

OR

$$f'(x) = 4x + k = 0$$     (M1)

Note: Award (M1) for setting the gradient function to zero.

$$k = – 5$$     (A1)     (C2)

(ii)     $$2{(1.25)^2} – 5(1.25) + 4$$     (M1)

$$= 0.875$$     (A1)(ft)     (C2)

Note: Follow through from their $$k$$.[4 marks]

a.

(A1)(ft)(A1)(ft)     (C2)

Notes: Award (A1)(ft) for a curve with correct concavity consistent with their $$k$$ passing through (0, 4).

(A1)(ft) for minimum in approximately the correct place. Follow through from their part (a).[2 marks]

b.

## Question

y = f (x) is a quadratic function. The graph of f (x) intersects the y-axis at the point A(0, 6) and the x-axis at the point B(1, 0). The vertex of the graph is at the point C(2, –2).

Write down the equation of the axis of symmetry.[2]

a.

Sketch the graph of y = f (x) on the axes below for 0 ≤ x ≤ 4 . Mark clearly on the sketch the points A , B , and C.

[3]

b.

The graph of y = f (x) intersects the x-axis for a second time at point D.

Write down the x-coordinate of point D.[1]

c.

## Markscheme

x = 2     (A1)(A1)     (C2)

Notes: Award (A1)(A0) for “ x = constant” (other than 2). Award (A0)(A1) for y = 2. Award (A0)(A0) for only seeing 2. Award (A0)(A0) for 2 = –b / 2a.[2 marks]

a.

(A1) for correctly plotting and labelling A, B and C

(A1) for a smooth curve passing through the three given points

(A1) for completing the symmetry of the curve over the domain given.     (A3)     (C3)

Notes: For A marks to be awarded for the curve, each segment must be a reasonable attempt at a continuous curve. If straight line segments are used, penalise once only in the last two marks.[3 marks]

b.

3     (A1)(ft)     (C1)

Notes: (A0) for coordinates. Accept x = 3 or D = 3 .[1 mark]

c.

## Question

Consider the two functions, $$f$$ and $$g$$, where

$$f(x) = \frac{5}{{{x^2} + 1}}$$

$$g(x) = {(x – 2)^2}$$

Sketch the graphs of $$y = f(x)$$ and $$y = g(x)$$ on the axes below. Indicate clearly the points where each graph intersects the y-axis.

[4]
a.

Use your graphic display calculator to solve $$f(x) = g(x)$$.[2]

b.

## Markscheme

$$f(x)$$: a smooth curve symmetrical about y-axis, $$f(x) > 0$$     (A1)

Note: If the graph crosses the x-axis award (A0).

Intercept at their numbered $$y = 5$$     (A1)

Note:  Accept clear scale marks instead of a number.

$$g(x)$$: a smooth parabola with axis of symmetry at about $$x = 2$$ (the 2 does not need to be numbered) and $$g(x) \geqslant 0$$     (A1)

Note: Right hand side must not be higher than the maximum of $$f(x)$$ at $$x = 4$$.

Accept the quadratic correctly drawn beyond $$x = 4$$.

Intercept at their numbered $$y = 4$$     (A1)     (C4)

Note: Accept clear scale marks instead of a number.[4 marks]

a.

$$–0.195, 2.76$$  $$(–0.194808…, 2.761377…)$$     (A1)(ft)(A1)(ft)     (C2)

Note: Award (A0)(A1)(ft) if both coordinates are given.

Follow through only if $$f(x) = \frac{5}{{{x^2}}} + 1$$ is sketched; the solutions are $$–0.841, 3.22$$  $$(–0.840913…, 3.217747…)$$[2 marks]

b.

## Question

Consider the functions $$f(x) = x + 1$$ and $$g(x) = {3^x} – 2$$.

Write down

(i)     the $$x$$-intercept of the graph of $$y = {\text{ }}f(x)$$;

(ii)     the $$y$$-intercept of the graph of $$y = {\text{ }}g(x)$$.[2]

a.

Solve $$f(x) = g(x)$$.[2]

b.

Write down the interval for the values of $$x$$ for which $$f(x) > g(x)$$.[2]

c.

## Markscheme

(i)     $$( – 1,{\text{ }}0)$$     (A1)

Note: Accept $$– 1$$.

(ii)     $$(0,{\text{ }} – 1)$$     (A1)     (C2)

Note: Accept $$– 1$$.

a.

$$(x = ){\text{ }} – 2.96{\text{ }}( – 2.96135 \ldots )$$     (A1)

$$(x = ){\text{ }}1.34{\text{ }}(1.33508 \ldots )$$     (A1)     (C2)

b.

$$– 2.96 < x < 1.34\;\;\;{\mathbf{OR}}\;\;\;\left] { – 2.96,{\text{ }}1.34} \right[\;\;\;{\mathbf{OR}}\;\;\;( – 2.96,{\text{ }}1.34)$$     (A1)(ft)(A1)     (C2)

Notes: Award (A1)(ft) for both correct endpoints of the interval, (A1) for correct strict inequalities (or correct open interval notation).

c.

## Question

The axis of symmetry of the graph of a quadratic function has the equation x $$= – \frac{1}{2}$$

Draw the axis of symmetry on the following axes.

The graph of the quadratic function intersects the x-axis at the point N(2, 0) . There is a second point, M, at which the graph of the quadratic function intersects the x-axis.[1]

a.

Draw the axis of symmetry on the following axes.

[1]

a.

The graph of the quadratic function intersects the $$x$$-axis at the point $${\text{N}}(2, 0)$$. There is a second point, $${\text{M}}$$, at which the graph of the quadratic function intersects the $$x$$-axis.

Clearly mark and label point $${\text{M}}$$ on the axes.[1]

b.

(i)     Find the value of $$b$$ and the value of $$c$$.

(ii)     Draw the graph of the function on the axes.[4]

c.

## Markscheme

vertical straight line which may be dotted passing through $$\left( { – \frac{1}{2},{\text{ }}0} \right)$$     (A1)     (C1)

a.

vertical straight line which may be dotted passing through $$\left( { – \frac{1}{2},{\text{ }}0} \right)$$     (A1)     (C1)

a.

point $${\text{M }}( – 3,{\text{ }}0)$$ correctly marked on the $$x$$-axis     (A1)(ft)     (C1)

Note: Follow through from part (a).

b.

(i)     $$b = 1$$, $$c = – 6$$     (A1)(ft)(A1)(ft)

(ii)     smooth parabola passing through $${\text{M}}$$ and $${\text{N}}$$     (A1)(ft)

Note: Follow through from their point $${\text{M}}$$ from part (b).

parabola passing through $$(0,{\text{ }} – 6)$$ and symmetrical about $$x = – 0.5$$     (A1)(ft) (C4)

Note: Follow through from part (c)(i).

If parabola is not smooth and not concave up award at most (A1)(A0).

c.

## Question

Consider the curve $$y = 1 + \frac{1}{{2x}},\,\,x \ne 0.$$

For this curve, write down

i)     the value of the $$x$$-intercept;

ii)    the equation of the vertical asymptote.[3]

a.

Sketch the curve for $$– 2 \leqslant x \leqslant 4$$ on the axes below.

[3]

b.

## Markscheme

i)     $$\left( {x = } \right) – 0.5\,\,\left( { – \frac{1}{2}} \right)$$       (A1)

ii)    $$x = 0$$        (A1)(A1)    (C3)

Note: Award (A1) for “$$x =$$” and (A1) for “$$0$$” seen as part of an equation.

a.

(A1)(ft)(A1)(ft)(A1)    (C3)

Note: Award (A1)(ft) for correct $$x$$-intercept, (A1)(ft) for asymptotic behaviour at $$y$$-axis, (A1) for approximately correct shape (cannot intersect the horizontal asymptote of $$y = 1$$). Follow through from part (a).

b.

## Question

The function $$f$$ is of the form $$f(x) = ax + b + \frac{c}{x}$$, where $$a$$ , $$b$$ and $$c$$ are positive integers.

Part of the graph of $$y = f(x)$$ is shown on the axes below. The graph of the function has its local maximum at $$( – 2,{\text{ }} – 2)$$ and its local minimum at $$(2,{\text{ }}6)$$.

Write down the domain of the function.[2]

a.

Draw the line $$y = – 6$$ on the axes.[1]

b.i.

Write down the number of solutions to $$f(x) = – 6$$.[1]

b.ii.

Find the range of values of $$k$$ for which $$f(x) = k$$ has no solution.[2]

c.

## Markscheme

$$(x \in \mathbb{R}),{\text{ }}x \ne 0$$     (A2)     (C2)

Note:     Accept equivalent notation. Award (A1)(A0) for $$y \ne 0$$.

Award (A1) for a clear statement that demonstrates understanding of the meaning of domain. For example, $${\text{D}}:( – \infty ,{\text{ }}0) \cup (1,{\text{ }}\infty )$$ should be awarded (A1)(A0).[2 marks]

a.

(A1)     (C1)

Note:     The command term “Draw” states: “A ruler (straight edge) should be used for straight lines”; do not accept a freehand $$y = – 6$$ line.[1 mark]

b.i.

2     (A1)(ft)     (C1)

Note:     Follow through from part (b)(i).[1 mark]

b.ii.

$$– 2 < k < 6$$     (A1)(A1)     (C2)

Note:     Award (A1) for both end points correct and (A1) for correct strict inequalities.

Award at most (A1)(A0) if the stated variable is different from $$k$$ or $$y$$ for example $$– 2 < x < 6$$ is (A1)(A0).[2 marks]

c.

## Question

The diagram shows part of the graph of a function $$y = f(x)$$. The graph passes through point $${\text{A}}(1,{\text{ }}3)$$.

The tangent to the graph of $$y = f(x)$$ at A has equation $$y = – 2x + 5$$. Let $$N$$ be the normal to the graph of $$y = f(x)$$ at A.

Write down the value of $$f(1)$$.[1]

a.

Find the equation of $$N$$. Give your answer in the form $$ax + by + d = 0$$ where $$a$$, $$b$$, $$d \in \mathbb{Z}$$.[3]

b.

Draw the line $$N$$ on the diagram above.[2]

c.

## Markscheme

3     (A1)     (C1)

Notes:     Accept $$y = 3$$[1 mark]

a.

$$3 = 0.5(1) + c$$$$\,\,\,$$OR$$\,\,\,$$$$y – 3 = 0.5(x – 1)$$     (A1)(A1)

Note:     Award (A1) for correct gradient, (A1) for correct substitution of $${\text{A}}(1,{\text{ }}3)$$ in the equation of line.

$$x – 2y + 5 = 0$$ or any integer multiple     (A1)(ft)     (C3)

Note:     Award (A1)(ft) for their equation correctly rearranged in the indicated form.

The candidate’s answer must be an equation for this mark.[3 marks]

b.

(M1)(A1)(ft)     (C2)

Note:     Award M1) for a straight line, with positive gradient, passing through $$(1,{\text{ }}3)$$, (A1)(ft) for line (or extension of their line) passing approximately through 2.5 or their intercept with the $$y$$-axis.[2 marks]

c.

## Question

Consider the graph of the function $$y = f(x)$$ defined below.

Write down all the labelled points on the curve

that are local maximum points;[1]

a.

where the function attains its least value;[1]

b.

where the function attains its greatest value;[1]

c.

where the gradient of the tangent to the curve is positive;[1]

d.

where $$f(x) > 0$$ and $$f'(x) < 0$$ .[2]

e.

B, F     (C1)

a.

H     (C1)

b.

F     (C1)

c.

A, E     (C1)

d.

C     (C2)

e.

## Question

On the grid below sketch the graph of the function $$f(x) = 2{(1.6)^x}$$ for the domain $$0 \leqslant x \leqslant 3$$ .

[2]

a.

Write down the coordinates of the $$y$$-intercept of the graph of $$y = f(x)$$ .[1]

b.

On the grid draw the graph of the function $$g(x) = 5 – 2x$$ for the domain $$0 \leqslant x \leqslant 3$$.[2]

c.

Use your graphic display calculator to solve $$f(x) = g(x)$$ .[1]

d.

## Markscheme

Note: Award (A1) correct endpoints, (A1) for smooth curve.     (A1)(A1)     (C2)

a.

$$(0{\text{, }}2)$$     (A1)     (C1)

Note: Accept $$x = 0$$, $$y = 2$$

b.

Straight line in the given domain     (A1)
Axes intercepts in the correct positions     (A1)     (C2)

c.

$$x = 0.943$$ ($$0.94259 \ldots$$)     (A1)     (C1)

Note: Award (A0) if $$y$$-coordinate given.

d.