Graphs and transformations: ACJC 2025 Promo Q6

(a)

(i) $y = -2\mathrm{f}(x) + 1$: asymptotes $y = 1$, $y = -5$ and $x = 1$; maximum point $(-1, 7)$; passes through $(0, 1)$ and $(2, 1)$.

(a)

(ii) $y = \dfrac{1}{\mathrm{f}(x)}$: asymptotes $x = 0$, $x = 2$ and $y = \tfrac{1}{3}$; maximum point $\left(-1, -\tfrac{1}{3}\right)$; passes through $(1, 0)$.

(a)

(iii) $y = \mathrm{f}(|x| + 1)$: asymptote $y = 3$ and $x = 0$; passes through $(-1, 0)$ and $(1, 0)$.

(b)

$y = 4x^2 - 4x + 1 = (2x - 1)^2$. Starting from $y = x^2$: translate $1$ unit in the positive $x$-direction to obtain $y = (x-1)^2$, then scale by scale factor $\tfrac{1}{2}$ parallel to the $x$-axis to obtain $y = (2x-1)^2$.