Transformations of Functions: past paper questions & worked solutions

The graph has amplitude $4$, crosses the origin, and initially moves downwards, so $f(x)=-4\sin x$.

The labelled maximum is at $P=\left(\dfrac{3\pi}{2},4\right)$.

$y=f\left(x-\dfrac\pi6\right)$ translates the graph right by $\dfrac\pi6$.

$P$ becomes $\left(\dfrac{3\pi}{2}+\dfrac\pi6,4\right)=\left(\dfrac{5\pi}{3},4\right)$.

$y=-\dfrac12f(x)$ leaves x-coordinates unchanged and multiplies y-coordinates by $-\dfrac12$.

$P$ becomes $\left(\dfrac{3\pi}{2},-2\right)$.

The original curve passes through $(-1,0)$ and $(0,2)$, and touches the x-axis at $(3,0)$.

$y=f(x+3)$ is a translation 3 units left. So $(-1,0)$ moves to $(-4,0)$, $(0,2)$ moves to $(-3,2)$, and $(3,0)$ moves to $(0,0)$.

The curve cuts the x-axis at $(-4,0)$ and touches it at $(0,0)$. The y-intercept is also $(0,0)$.

For $y=f(-3x)$, a point with original x-coordinate $X$ moves to $x=-\dfrac{X}{3}$.

$(-1,0)$ moves to $(\dfrac13,0)$, $(3,0)$ moves to $(-1,0)$, and $(0,2)$ stays at $(0,2)$.

Subtracting 2 moves every point of the graph down by 2 units.

The minimum $P(-1,0)$ becomes $(-1,-2)$, the maximum $Q(\frac32,2)$ becomes $(\frac32,0)$, and the asymptote $y=1$ becomes $y=-1$.

Replacing $x$ by $-x$ reflects the graph in the y-axis.

The minimum $P(-1,0)$ becomes $(1,0)$, the maximum $Q(\frac32,2)$ becomes $(-\frac32,2)$, and the horizontal asymptote remains $y=1$.

The curve is below the line $y=6$ for $x<-4$, for $-1<x<2$, and for $x>3$.

Since $f(u)=6$ at $u=-4,-1,2,3$, we set $2x=-4,-1,2,3$. The solutions are $x=-2,-\dfrac12,1,\dfrac32$.

The largest solution is $x=\dfrac32$.

For $y=f(-x)$, each point $(x,y)$ becomes $(-x,y)$.

$P(-4,6)\to(4,6)$, $Q(-1,6)\to(1,6)$, $R(2,6)\to(-2,6)$, and $S(3,6)\to(-3,6)$.

For $x<0$, the part of the reflected graph on or above $y=6$ is from $x=-3$ to $x=-2$.

$y=f(2x)$ is a horizontal scale factor of $\dfrac12$, so $P(9,3)$ maps to $\left(\dfrac92,3\right)$.

$y=f(x)+3$ is a translation 3 units upward, so $P(9,3)$ maps to $(9,6)$.

At $Q$, $f(2x)=f(x)+3$. Since $f(x)=\sqrt{x}$, this gives $\sqrt{2x}=\sqrt{x}+3$.

Let $s=\sqrt{x}$. Then $\sqrt{2x}=\sqrt2\,s$, so $\sqrt2s=s+3$. Hence $(\sqrt2-1)s=3$ and $s=\dfrac3{\sqrt2-1}=3(\sqrt2+1)$.

$\sqrt{x}=3(\sqrt2+1)$, so $x=9(\sqrt2+1)^2=9(3+2\sqrt2)=27+18\sqrt2$.

$y=\sqrt{x}+3=3(\sqrt2+1)+3=6+3\sqrt2$. Therefore $Q=(27+18\sqrt2,\ 6+3\sqrt2)$.

$C_1$ has equation $y=\sqrt{2x}$. This is obtained from $y=\sqrt{x}$ by a stretch parallel to the x-axis with scale factor $\dfrac12$.

Equivalently, $y=\sqrt{2x}=\sqrt2\sqrt{x}$ is a stretch parallel to the y-axis with scale factor $\sqrt2$. The mark scheme also accepts the horizontal stretch description.

$C_2$ has equation $y=12-\sqrt{x}=-\sqrt{x}+12$, so it is a translation of $y=-\sqrt{x}$ by $\begin{pmatrix}0\\12\end{pmatrix}$.

At $P$, $\sqrt{2x}=12-\sqrt{x}$.

$\sqrt2\sqrt{x}+\sqrt{x}=12$, so $(\sqrt2+1)\sqrt{x}=12$.

$\sqrt{x}=\dfrac{12}{\sqrt2+1}=\dfrac{12(\sqrt2-1)}{(\sqrt2+1)(\sqrt2-1)}=12(\sqrt2-1)$.

$x=[12(\sqrt2-1)]^2=144(3-2\sqrt2)$.

$y=\sqrt{2x}=\sqrt2\sqrt{x}=\sqrt2\cdot12(\sqrt2-1)=24-12\sqrt2$.

The graph is a rectangular hyperbola with asymptotes $x=0$ and $y=0$, with branches in quadrants I and III.

$y=\dfrac6{x-2}$ is obtained from $y=\dfrac6x$ by a translation 2 units in the positive x-direction.

At $x=-4$ on $y=\dfrac6{x-2}$, $y=\dfrac6{-6}=-1$.

Since $P$ lies on $y=kx+7$, $-1=-4k+7$, so $k=2$.

Solve $\dfrac6{x-2}=2x+7$. Multiplying by $x-2$ gives $6=(2x+7)(x-2)$.

$6=2x^2+3x-14$, so $2x^2+3x-20=0=(2x-5)(x+4)$. The root $x=-4$ is $P$, so $Q$ has $x=\dfrac52$.

$y=2\left(\dfrac52\right)+7=12$, so $Q=\left(\dfrac52,12\right)$.