Curve sketching, modulus graphs and transformations – A-level Maths exam questions with mark scheme

1StandardCalculator6 marks

(a) Sketch the graph of $y=|2x-3|$, stating the coordinates of any points where the graph meets or cuts the coordinate axes. [2]

(b) Solve the equation $$|2x-3|=5-x$$ [4]

(a) V-shaped graph with its vertex on the positive $x$-axis (both branches above the $x$-axis)B1
Both $\left(\tfrac32,0\right)$ and $(0,3)$ correctly stated or marked on the sketchB1
(b) Solves $2x-3=5-x$ to find a value of $x$M1
$x=\dfrac{8}{3}$A1
Solves $-(2x-3)=5-x$ (or $3-2x=5-x$) to find a value of $x$M1
$x=-2$ (both $x=\tfrac83$ and $x=-2$ given, and no other values)A1
Answer: (a) V-shape, vertex $(1.5,0)$, $y$-intercept $(0,3)$ (b) $x=\tfrac83$ or $x=-2$
Examiner tip: The second solution comes from the reflected branch, $-(2x-3)=5-x$ – drawing $y=5-x$ on your sketch shows there are two intersections.

2HardCalculator7 marks

The diagram shows a sketch of the curve with equation $y=\mathrm{f}(x)$, $x\in\mathbb{R}$. The curve has a maximum turning point at $P(-2,6)$ and a minimum turning point at $Q(3,-4)$. The curve has no other turning points and continues indefinitely in both directions as shown.

(a) State the coordinates of the stationary points on the graph of $y=|\mathrm{f}(x)|$. [2]

(b) State the coordinates of the points to which $P$ and $Q$ are mapped on the curve with equation $y=2\mathrm{f}(x-1)+3$. [3]

(a) $(-2,6)$B1
$(3,4)$ (the cusps where the graph meets the $x$-axis are not stationary points and are not required; do not award if extra stationary points are given)B1
(b) Either $x$-coordinate correct: $-1$ or $4$ (horizontal translation of $+1$)B1
$P\mapsto(-1,15)$B1
$Q\mapsto(4,-5)$B1
(c) Uses the $y$-coordinates of the turning points as the critical values, $-4$ and $6$M1
$-4A1
Answer: (a) $(-2,6)$ and $(3,4)$ (b) $(-1,15)$ and $(4,-5)$ (c) $-4
Examiner tip: In $2\mathrm{f}(x-1)+3$ the 2 is applied before the $+3$: $y\mapsto 2y+3$, so $-4\mapsto-5$, not $-2$.

3StretchCalculator8 marks

Given that $a$ is a positive constant, the function $\mathrm{f}$ is defined by $$\mathrm{f}(x)=2|x+a|-3a,\quad x\in\mathbb{R}$$

(a) Sketch the graph of $y=\mathrm{f}(x)$, showing the coordinates of the vertex and of any points where the graph meets the coordinate axes, in terms of $a$. [3]

(b) Solve, in terms of $a$, the equation $\mathrm{f}(x)=x+a$. [4]

(a) V-shape with vertex at $(-a,-3a)$ in the third quadrantB1
$y$-intercept $(0,-a)$B1
$x$-intercepts $\left(-\tfrac52a,0\right)$ and $\left(\tfrac12a,0\right)$B1
(b) Solves $2(x+a)-3a=x+a$ to find $x$M1
$x=2a$A1
Solves $-2(x+a)-3a=x+a$ to find $x$M1
$x=-2a$ (both correct and no others)A1
(c) $k<-3a$B1
Answer: (a) vertex $(-a,-3a)$; meets axes at $(0,-a)$, $\left(-\tfrac52a,0\right)$, $\left(\tfrac12a,0\right)$ (b) $x=2a$ or $x=-2a$ (c) $k<-3a$
Examiner tip: Check each case actually lies on the line – with a parameter it is easy to keep an extra root that does not satisfy the original modulus equation.

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4HardCalculator8 marks

The diagram shows a sketch of the curve $C$ with equation $y=(x+1)(x-3)^{2}$ and the straight line $l$ with equation $y=9-3x$.

(a) Write down the coordinates of the points where $C$ meets the coordinate axes. [2]

(b) Using algebra, find the coordinates of all the points where $l$ meets $C$. [4]

The curve with equation $y=f(x-a)$, where $f(x)=(x+1)(x-3)^{2}$ and $a$ is a constant, passes through the origin.

(a) $(-1,0)$ and $(3,0)$B1
(a) $(0,9)$B1
(b) for setting $(x+1)(x-3)^{2}=9-3x$M1
(b) for collecting into a cubic $=0$ and factorising, e.g. $x^{3}-5x^{2}+6x=0$ or $(x-3)\left[(x+1)(x-3)+3\right]=0$M1
(b) $x=0,\ 2,\ 3$A1
(b) $(0,9),\ (2,3),\ (3,0)$A1
(c) one correct value, $a=1$ or $a=-3$B1
(c) both $a=1$ and $a=-3$ (and no others)B1
Answer: (a) $(-1,0),(3,0),(0,9)$ (b) $(0,9),(2,3),(3,0)$ (c) $a=1$ or $a=-3$
Examiner tip: Don't divide both sides by $(x-3)$ in (b) without recording $x=3$ as a solution – you'd lose a point.

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