Graphs, key features, solving graphically and transformations – IB Maths exam questions with mark scheme

1StandardNon-calculator5 marks

The point $A(4,-2)$ lies on the graph of $y=f(x)$. Find the coordinates of the image of $A$ on the graph of

(a) $y=f(x-3)+1$ [1]

(b) $y=-2f(x)$ [1]

(d) $y=3f(2x-4)$ [2]

$(7,-1)$A1
$(4,4)$A1
$(2,-2)$A1
Solves $2x-4=4$ so $x=4$M1
$(4,-6)$A1
Answer: (a) $(7,-1)$ (b) $(4,4)$ (c) $(2,-2)$ (d) $(4,-6)$
Examiner tip: Horizontal changes act on $x$ in the opposite sense to how they look; solve inside the bracket equals 4.

2CoreNon-calculator5 marks

The graph of $y=f(x)$ has a vertical asymptote $x=1$ and a horizontal asymptote $y=2$.

The function $g$ is defined by $g(x)=2f(x+3)-1$.

(a) Describe fully a sequence of three transformations that maps the graph of $y=f(x)$ onto the graph of $y=g(x)$. [3]

(b) Write down the equation of

(i) the vertical asymptote of the graph of $y=g(x)$;

(ii) the horizontal asymptote of the graph of $y=g(x)$. [2]

(a) Horizontal translation of $3$ units to the left (accept translation by $\begin{pmatrix}-3\\0\end{pmatrix}$); may appear at any point in the sequenceA1
Vertical stretch, scale factor $2$ (accept 'parallel to the $y$-axis', 'in the $y$-direction'; do not accept 'enlargement' or 'stretch' with no direction)A1
Vertical translation of $1$ unit down (accept $\begin{pmatrix}0\\-1\end{pmatrix}$), applied after the vertical stretch. Accept 'translate down $\frac12$ then stretch vertically by factor 2'. Award A0 if the downward translation by 1 is stated before the stretchA1
(b)(i) $x=-2$A1
(ii) $y=3$A1
Answer: (a) Translate 3 left; vertical stretch sf 2; then translate 1 down (b)(i) $x=-2$ (ii) $y=3$
Examiner tip: The order matters for the vertical changes: stretch by 2 and then move down 1, otherwise you get $2(f-1)=2f-2$, not $2f-1$.

3CoreCalculator7 marks

Let $f(x)=x^3-3x^2-x+2$, for $-1\le x\le 4$.

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

(b) Find the zeros of $f$. [2]

(i) the local maximum point;

(ii) the local minimum point. [2]

(d) Hence find the range of $f$. [2]

(a) $2$ (accept $(0,2)$)A1
(b) Any two of $x=-0.861,\ 0.746,\ 3.11$A1
All three correct and no extra values. Award A1A0 for answers given only to 2 s.f. throughoutA1
(c)(i) $(-0.155,\ 2.08)$A1
(ii) $(2.15,\ -4.08)$A1
(d) Evaluates end point: $f(4)=14$ (may be seen in range). Must consider end points, since $f(-1)=-1$ and the local maximum $2.08$ is not the greatest valueM1
$-4.08\le f(x)\le 14$ (accept $[-4.08,\,14]$; accept $y$ in place of $f(x)$; A0 for strict inequalities)A1
Answer: (a) $2$ (b) $x=-0.861,\ 0.746,\ 3.11$ (c)(i) $(-0.155,\,2.08)$ (ii) $(2.15,\,-4.08)$ (d) $-4.08\le f(x)\le14$
Examiner tip: The range on a restricted domain comes from the end points as well as the turning points: here $f(4)=14$, far above the local maximum.

Stuck on Graphs, key features, solving graphically and transformations?

Get it sorted in a free 30-minute session

We work through your weakest questions on this topic on Zoom. I reply the same day.

Free 30-min session on this topic Send me 5 questions a day

4StandardCalculator5 marks

Consider the equation $2^x=x^2$.

(a) Show that $x=2$ and $x=4$ are both solutions of the equation. [1]

(b) By sketching the graphs of $y=2^x$ and $y=x^2$ on your graphic display calculator, find the other solution of the equation. [2]

(a) Substitutes both: $2^2=4=2^2$ and $2^4=16=4^2$ (both needed)A1
(b) Valid graphical approach, e.g. sketch of both curves showing an intersection for $x<0$, or uses $y=2^x-x^2$ and finds its zerosM1
$x=-0.767$ (accept $-0.76666\ldots$)A1
(c) Recognises that $x+1$ takes the values $-0.767,\ 2,\ 4$ (or a translation of $1$ unit left), e.g. $x+1=2$M1
$x=-1.77,\ 1,\ 3$ (all three; follow through from their (b))A1
Answer: (b) $x=-0.767$ (c) $x=-1.77,\ 1,\ 3$
Examiner tip: Many students only find the two obvious integer solutions: an exponential and a quadratic can meet three times, so look at negative $x$ on the GDC.

5StandardNon-calculator8 marks

The diagram shows the graph of $y=f(x)$ for $-4\le x\le 4$. The graph consists of three straight line segments joining the points $(-4,2)$, $(-2,-2)$, $(1,4)$ and $(4,4)$.

(a) Solve $f(x)=0$. [2]

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

The function $g$ is defined by $g(x)=f(2x)-3$.

(d) Write down the coordinates of the minimum point of the graph of $y=g(x)$. [1]

(e) Find the set of values of $k$ for which the equation $f(x)=k$ has exactly two solutions. [2]

(a) Either correct zero found, e.g. via line equation $y=-2x-6$ or $y=2x+2$, or read from the graph: $x=-3$ or $x=-1$A1
Both $x=-3$ and $x=-1$, and no othersA1
(b) $-2\le f(x)\le 4$ (accept $[-2,4]$)A1
(c) Domain $-2\le x\le 2$A1
Range $-5\le g(x)\le 1$A1
(d) $(-1,\,-5)$A1
(e) Considers a horizontal line $y=k$ crossing both the first and second segments, e.g. identifies $k=2$ (from $(-4,2)$) or $k=-2$ as boundary valuesM1
$-2<k\le 2$ (A0 for $-2<k<2$ or $-2\le k\le 2$)A1
Answer: (a) $x=-3,\ -1$ (b) $-2\le f(x)\le4$ (c) domain $-2\le x\le2$, range $-5\le g(x)\le1$ (d) $(-1,-5)$ (e) $-2<k\le2$
Examiner tip: In (e) check the end values: $k=2$ still gives two solutions ($x=-4$ and $x=0$), while $k=-2$ gives only one and $k=4$ gives infinitely many.

6StandardCalculator7 marks

The concentration of a drug in a patient's bloodstream, $C$ mg per litre, $t$ hours after the drug is taken is modelled by

$$C(t)=20t\,e^{-0.4t},\quad t\ge 0.$$

(a) Find the maximum concentration and the time at which it occurs. [2]

(b) The drug is effective when the concentration is at least $15$ mg per litre. Find the length of time for which the drug is effective. [3]

(ii) Interpret this asymptote in the context of the model. [2]

(a) $t=2.5$ (hours)A1
$C=18.4$ mg per litre (accept $18.39\ldots$)A1
(b) Attempt to solve $20te^{-0.4t}=15$, e.g. sketch with the line $C=15$ drawn, or uses intersect on GDCM1
$t=1.22$ and $t=4.45$ (both)A1
$3.23$ hours (accept $3.22984\ldots$; accept $3$ h $14$ min)A1
(c)(i) $C=0$ (accept $y=0$)A1
(ii) In the long run the drug is (almost) completely removed from the bloodstream / the concentration tends to zero but never reaches exactly zeroR1
Answer: (a) $18.4$ mg/L at $t=2.5$ h (b) $3.23$ hours (c)(i) $C=0$ (ii) concentration tends to zero over time
Examiner tip: Students often give the two times from (b) as the answer; the question asks for the length of time, so subtract them.

7HardNon-calculator7 marks

Let $f(x)=x^2-4x$.

The graph of $y=f(x)$ is transformed to the graph of $y=g(x)$ by the following two transformations, applied in this order:

$T_1$: a horizontal stretch with scale factor $\frac12$;

$T_2$: a translation by the vector $\begin{pmatrix}3\\-2\end{pmatrix}$.

(a) Show that $g(x)=4x^2-32x+58$. [3]

The transformations are now applied in the opposite order, $T_2$ then $T_1$, to the graph of $y=f(x)$, giving the graph of $y=h(x)$.

(b) Find $h(x)$, giving your answer in the form $ax^2+bx+c$. [2]

(c) The vertex of the graph of $g$ is $(4,-6)$. Find the vertex of the graph of $h$, and hence describe the single transformation that maps the graph of $y=g(x)$ onto the graph of $y=h(x)$. [2]

(a) After $T_1$: $f(2x)$ seenA1
Replaces $x$ by $x-3$ inside and subtracts 2: $g(x)=f(2(x-3))-2$ or $(2x-6)^2-4(2x-6)-2$M1
Correct expansion $4x^2-24x+36-8x+24-2$ leading to $4x^2-32x+58$ (AG: must see an intermediate expanded line)A1
(b) $h(x)=f(2x-3)-2$ or $(2x-3)^2-4(2x-3)-2$ seenM1
$h(x)=4x^2-20x+19$A1
(c) Vertex of $h$: $\left(\frac52,\,-6\right)$ (accept $(2.5,-6)$)A1
Translation by $\begin{pmatrix}-\frac32\\0\end{pmatrix}$ (accept 'translation $\frac32$ units to the left'; follow through from their vertex)A1
Answer: (b) $h(x)=4x^2-20x+19$ (c) vertex $(2.5,-6)$; translation by $\begin{pmatrix}-1.5\\0\end{pmatrix}$
Examiner tip: When a horizontal stretch comes after a horizontal translation, the translation is also stretched: $f(2x-3)$ shifts the graph by $1.5$, not $3$.

8HardCalculator8 marks

Let $f(x)=(x^2-1)e^{-x}$, for $x\in\mathbb{R}$.

(a) Find the zeros of $f$. [2]

(b) Find the coordinates of the local minimum point and the local maximum point of the graph of $f$. [2]

(d) Find the set of values of $k$ for which the equation $f(x)=k$ has exactly three solutions. Justify your answer. [3]

(a) $x=1$ or $x=-1$ (one correct)A1
Both $x=\pm1$ and no othersA1
(b) Local minimum $(-0.414,\,-1.25)$A1
Local maximum $(2.41,\,0.432)$ (accept exact $x=1\pm\sqrt2$)A1
(c) $y=0$A1
(d) Considers horizontal lines $y=k$ meeting the graph, e.g. sketch or reference to the local maximum value and the asymptoteM1
Lower bound: $k>0$, with reason, e.g. the graph approaches but never reaches $y=0$ as $x\to\infty$ (so $k=0$ gives only two solutions)R1
$0<k<0.432$ (A0 if either end point included)A1
Answer: (a) $x=\pm1$ (b) min $(-0.414,-1.25)$, max $(2.41,0.432)$ (c) $y=0$ (d) $0<k<0.432$
Examiner tip: The lower bound comes from the asymptote, not the minimum: for $-1.25<k\le0$ the right-hand branch never reaches the line, so there are only two solutions.

9StretchNon-calculator9 marks

Let $f(x)=\ln x$ and $g(x)=\ln\left(e^2x^3\right)$, for $x>0$.

(a) Show that $g(x)=3f(x)+2$. [2]

(b) Hence describe a sequence of two transformations that maps the graph of $y=f(x)$ onto the graph of $y=g(x)$. [2]

Amara claims that the graph of $y=g(x)$ can also be obtained from the graph of $y=f(x)$ by a horizontal stretch followed by a vertical stretch with scale factor $3$, with no translation.

(d) The point $P(e,\,1)$ lies on the graph of $y=f(x)$. Find the image of $P$ under each sequence of transformations, and verify that both images lie on the graph of $y=g(x)$. [2]

(a) Uses $\ln(ab)=\ln a+\ln b$: $\ln e^2+\ln x^3$M1
$2+3\ln x$, so $g(x)=3f(x)+2$ (AG)A1
(b) Vertical stretch, scale factor $3$A1
followed by a translation by $\begin{pmatrix}0\\2\end{pmatrix}$ (accept: translate up $\frac23$ then stretch vertically by 3)A1
(c) Writes $e^2x^3$ as a cube, e.g. $e^2x^3=\left(e^{2/3}x\right)^3$M1
$g(x)=3\ln\left(e^{2/3}x\right)=3f\left(e^{2/3}x\right)$A1
Horizontal stretch, scale factor $e^{-2/3}$ (accept $\frac{1}{e^{2/3}}$; A0 for $e^{2/3}$)A1
(d) Images $(e,\,5)$ and $\left(e^{1/3},\,3\right)$A1
Verifies $g(e)=\ln e^5=5$ and $g\left(e^{1/3}\right)=\ln e^3=3$, so both lie on the curve (the two routes give the same curve but move individual points differently)R1
Answer: (b) vertical stretch sf 3, then translate up 2 (c) horizontal stretch sf $e^{-2/3}$ (d) $(e,5)$ and $(e^{1/3},3)$
Examiner tip: For $f(ax)$ the horizontal scale factor is $\frac1a$, so $3f(e^{2/3}x)$ needs a factor of $e^{-2/3}$, not $e^{2/3}$.

10StandardCalculator9 marks

A cup of coffee is left to cool in a room. Its temperature, $T\,^{\circ}\mathrm{C}$, $t$ minutes after it is poured is modelled by

$$T(t)=20+70e^{-0.05t},\quad 0\le t\le 60.$$

(a) Write down the temperature of the coffee when it is poured. [1]

(b) (i) Write down the equation of the horizontal asymptote of the graph of $T$ if the domain were extended to $t\ge0$.

(ii) Interpret this value in context. [2]

(d) Sketch the graph of $T$ for $0\le t\le 60$, labelling the coordinates of its end points. [3]

A second cup is poured from the same pot $5$ minutes later. Its temperature is given by $S(t)=T(t-5)$ for $5\le t\le 65$.

(e) Describe the transformation that maps the graph of $T$ onto the graph of $S$. [1]

(a) $90\,^{\circ}\mathrm{C}$A1
(b)(i) $T=20$ (accept $y=20$)A1
(ii) The temperature of the room / the coffee cools towards $20\,^{\circ}\mathrm{C}$ but never falls below itR1
(c) Sets $20+70e^{-0.05t}=40$ (or uses GDC intersect with $T=40$)M1
$t=25.1$ minutes (accept $25.0552\ldots$; exact $20\ln3.5$)A1
(d) Decreasing curve, concave up (gradient becoming less steep), staying above $T=20$A1
Starts at $(0,\,90)$ labelledA1
Ends at $(60,\,23.5)$ labelled (accept $23.49$; condone a point not at exact scale if labelled)A1
(e) Translation by $\begin{pmatrix}5\\0\end{pmatrix}$ (accept '5 units to the right' / 'in the positive $t$ direction'; A0 for 'left')A1
Answer: (a) $90\,^{\circ}\mathrm{C}$ (b) $T=20$, room temperature (c) $25.1$ min (d) decreasing concave-up curve from $(0,90)$ to $(60,23.5)$ (e) translation 5 to the right
Examiner tip: In (d) the sketch must show the curve flattening towards $20$, not a straight line, and the end points must be labelled to earn all three marks.

Stuck on this topic?

Book a free 30-minute session on WhatsApp and we'll find exactly which step is costing you the marks. Or build a timed paper on it.

Free 30-minute session