Graphs and transformations: ACJC 2025 Promo Q6

Anglo-Chinese Junior College2025 PromoPromo Q6●●● Challenging11 marks

What this question tests

Graphing Transformations.

Question

The figure below shows the sketch of the graph $y = \mathrm{f}(x)$ . The graph has asymptotes at $x = 1$ , $y = 3$ and the $x$ -axis. The graph has a stationary point at $(-1, -3)$ and cuts the axes at $(0, 0)$ and $(2, 0)$ .
$Anglo-Chinese Junior College 2025 Promo Q6: question diagram$
Sketch, on separate diagrams, the following graphs, indicating clearly any asymptotes, axial intercepts and stationary points, where possible.
1. $y = -2\mathrm{f}(x) + 1$ [3]
2. $y = \dfrac{1}{\mathrm{f}(x)}$ [3]
3. $y = \mathrm{f}(|x| + 1)$ [3]
Describe fully a sequence of transformations that transform the graph of $y = x^2$ onto the graph of $y = 4x^2 - 4x + 1$ . [2]

Show full worked solution▾

(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 $(2, 1)$ .

$Anglo-Chinese Junior College 2025 Promo Q6: solution diagram$

(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 $(1, 0)$ .

$Anglo-Chinese Junior College 2025 Promo Q6: solution diagram$

(a)

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

$Anglo-Chinese Junior College 2025 Promo Q6: solution diagram$

(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 $y = (2x-1)^2$ .

Answers

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

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