<ol class="abc-list"><li><div class="list-item-paragraph">The graph of $y=\mathrm{f}(x)$ is shown in the diagram. It passes through the origin and has a minimum point at $\left( 2,\mathrm{ }0 \right)$. The lines $x=1$ and $y=2$ are the asymptotes of the graph.</div><div class="content-block list-item-image" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/public/question-images/converted/1759755650243-f34pm5sc6li.png" alt="List item image" class="aligncenter custom-width" style="max-width: min(450px, 100%); height: auto;" /></div><div class="list-item-paragraph">Sketch the graph of $y=2-\mathrm{f}(x)$, stating the coordinates of the point where the graph crosses the $y$-axis, the coordinates of any stationary point(s) and the equations of any asymptotes.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">The curve $C$ has equation $y=\frac{2{{x}^{2}}+kx-2}{x+2}$, where&nbsp;$k$ is a constant.</div><ol class="i-list"><li><div class="list-item-paragraph">Find the set of values of $k$ for which $C$ has&nbsp;$2$ stationary points.<span class="inline-marks">[3]</span></div><div class="list-item-paragraph">For the rest of the question, take $k=-3$.</div></li><li><div class="list-item-paragraph">By expressing the equation $y=\frac{2{{x}^{2}}-3x-2}{x+2}$ in the form $y=ax+b+\frac{c}{x+2}$, where&nbsp;$a$, $b$ and $c$ are constants, write down the equations of the asymptotes of $C$.<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">Hence sketch $C$, giving the equations of any asymptotes, the coordinates of any stationary points and of the points where&nbsp;$C$ crosses the axes.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">Describe a pair of transformations which transform the graph of $y=x+\frac{6}{x+2}$ to $C$.<span class="inline-marks">[2]</span></div></li></ol></li></ol>

The graph of $y=\mathrm{f}(x)$ is shown in the diagram. It passes through the origin and has a minimum point at $\left( 2,\mathrm{ }0 \right)$. The lines $x=1$ and $y=2$ are the asymptotes of the graph.

Sketch the graph of $y=2-\mathrm{f}(x)$, stating the coordinates of the point where the graph crosses the $y$-axis, the coordinates of any stationary point(s) and the equations of any asymptotes.

The curve $C$ has equation $y=\frac{2{{x}^{2}}+kx-2}{x+2}$, where $k$ is a constant.

Transformations of Graphs - General

Transformations of Graphs

International Baccalaureate Higher Level Mathematics

IB HL Math

2020 VJC Promo Q8

(a) $y = \text{f}(x) \to y = 2 - \text{f}(x)$

Step A: Substitute $y$ as $-y$ → Reflect about the x-axis.

$y = -\text{f}(x)$

Step B: Substitute $y$ as $y - 2$ → Translation of 2 units in the positive y-direction.

$y = 2 - \text{f}(x)$

Graph of $y = 2 - \text{f}(x)$:

Key features: Asymptotes $x = 1$ (VA), $y = 0$ (HA becomes $y = 0$ after reflection and shift)

Key points: $(0, 2)$, $(2, 2)$

(b)(i) $y = \frac{2x^2 + kx - 2}{x + 2}$

$\frac{dy}{dx} = \frac{(x+2)(4x+k) - (2x^2+kx-2)(1)}{(x+2)^2}$

$= \frac{4x^2 + kx + 8x + 2k - 2x^2 - kx + 2}{(x+2)^2}$

$= \frac{2x^2 + 8x + 2 + 2k}{(x+2)^2}$

For C to have 2 stationary values, $\frac{dy}{dx} = 0$ has 2 real solutions.

$2x^2 + 8x + 2 + 2k = 0$

Discriminant $D > 0$:

$8^2 - 4(2)(2 + 2k) > 0$

$64 - 8(2 + 2k) > 0$

$8 - 2 - 2k > 0$

$6 > 2k$

$3 > k$

Solution set: $\{k \in \mathbb{R} : k < 3\}$

(b)(ii) By long division:

$\frac{2x^2 - 3x - 2}{x + 2} = 2x - 7 + \frac{12}{x + 2}$

Oblique asymptote: $y = 2x - 7$

Vertical asymptote: $x = -2$

Key points: $\left(-\frac{1}{2}, 0\right)$, $(2, 0)$, $(0, -1)$

Stationary points: $(0.449, -1.20)$ and $(-4.45, -20.8)$

(b)(iv) $y = x + \frac{6}{x+2} \to y = 2x - 7 + \frac{12}{x+2}$

Sequence of transformations:

I: Scaling by a factor of 2 parallel to the y-axis (replace $y$ with $\frac{y}{2}$).

$2y = 2x + \frac{12}{x+2}$

II: Translation of 7 units in the negative y-direction (replace $y$ with $y + 7$).

$y = 2x + \frac{12}{x+2} - 7 = 2x - 7 + \frac{12}{x+2}$

2020 VJC Promo Q8

2020 VJC Promo Q8

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