2015 H2 Maths Paper 1, Question 5

We first replace $x$ by $x - 3$, which translates the curve $3$ units in the positive $x$-direction:

We then multiply the whole expression by $\frac{1}{4}$, which scales the curve parallel to the $y$-axis by factor $\frac{1}{4}$:

On $0 \leq x \leq 1$, $f(x) = 1$, a horizontal segment from $(0,1)$ to $(1,1)$, both endpoints attained.

On $1 < x \leq 3$, $f(x) = \frac{1}{4}(x - 3)^{2}$, part of an upward parabola with vertex $(3,0)$. At $x = 1$ (not included in this piece, but the value matches the flat segment already found):

So the curve comes in at $(1,1)$ with a corner (the gradient jumps from $0$ to $- 1$ there), then decreases to $(3,0)$.

Otherwise, $f(x) = 0$. For $- 1 \leq x < 0$ this gives a horizontal segment on the $x$-axis, but at $x = 0$ itself $f(0) = 1$ (the first piece applies), so there is a jump: $f$ approaches $0$ as $x \rightarrow 0^{-}$ but $f(0) = 1$. For $3 < x \leq 4$, $f(x) = 0$, and since $f(3) = 0$ too, the curve meets this segment smoothly (no jump) at $(3,0)$.

Note: the open circle at $(0,0)$ marks the value approached but not attained; $(0,1)$ is the actual value of $f(0)$.

For $0 \leq \frac{1}{2}x \leq 1$, i.e. $0 \leq x \leq 2$: $f\left( \frac{1}{2}x \right) = 1$, so $y = 2$.

For $1 < \frac{1}{2}x \leq 3$, i.e. $2 < x \leq 6$: $f\left( \frac{1}{2}x \right) = \frac{1}{4}\left( \frac{1}{2}x - 3 \right)^{2}$, so

At $x = 4$ (the right end of our domain):

Otherwise (in particular for $- 1 \leq x < 0$): $f\left( \frac{1}{2}x \right) = 0$, so $y = 1$; but at $x = 0$ itself, $y = 1 + f(0) = 2$. As in part (ii), this is a jump: $y \rightarrow 1$ as $x \rightarrow 0^{-}$, but $y(0) = 2$.