<p>It is given that $y=\frac{2{{x}^{2}}+3}{x-2},x\in \mathbb{R},x\ne 2$.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Sketch the graph of $y=\frac{2{{x}^{2}}+3}{x-2}$ and label clearly the equations of the asymptotes and intercepts if any.</div></li><li><div class="list-item-paragraph">The graph of $y=\frac{2{{x}^{2}}+3}{x-2}$ cannot lie between values $p$ and $m$, where $p>m$ . State the value of $p$ and $m$ .</div></li><li><div class="list-item-paragraph">By means of a graphical argument, state the maximum number of real roots the equation $kx\left( x-2 \right)-2{{x}^{2}}=3$ have.</div></li></ol>

It is given that $y=\frac{2{{x}^{2}}+3}{x-2},x\in \mathbb{R},x\ne 2$.

Sketch the graph of $y=\frac{2{{x}^{2}}+3}{x-2}$ and label clearly the equations of the asymptotes and intercepts if any.

The graph of $y=\frac{2{{x}^{2}}+3}{x-2}$ cannot lie between values $p$ and $m$, where $p>m$ . State the value of $p$ and $m$ .

By means of a graphical argument, state the maximum number of real roots the equation $kx\left( x-2 \right)-2{{x}^{2}}=3$ have.

Parametric Equations

Graphing Techniques

International Baccalaureate Higher Level Mathematics

IB HL Math

ACJC Tutorial 2 Q8

(i) By long division:

$\frac{2x^2 + 3}{x - 2} = 2x + 4 + \frac{11}{x-2}$

OA: $y = 2x + 4$

VA: $x = 2$

Key points: $(0, -\frac{3}{2})$, stationary points at $(0 + 2\sqrt{22}, 8 + 2\sqrt{22})$ and similar.

Graph sketch with curve approaching asymptotes.

(ii) For $y = k$, where $k \in \mathbb{R}$, the horizontal line $y = k$ cuts the curve.

For the line $y = k$ to be tangent to the curve:

$k = 2x + 4 + \frac{11}{x-2}$

$k(x-2) = 2x^2 + 3$

$2x^2 - kx + 2k + 3 = 0$

For no real solutions (line doesn't intersect):

$k^2 - 4(2)(2k+3) < 0$ (wait, this is for tangent)

Actually: $k^2 - 8(2k+3) < 0$

$k^2 - 16k - 24 < 0$

$k = \frac{16 \pm \sqrt{256 + 96}}{2} = \frac{16 \pm \sqrt{352}}{2} = \frac{16 \pm 4\sqrt{22}}{2} = 8 \pm 2\sqrt{22}$

$\therefore p = 8 + 2\sqrt{22}$, $m = 8 - 2\sqrt{22}$

(iii) $kx(x-2) - 2x^2 = 3$

$kx(x-2) = 2x^2 + 3$

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

Since the line $y = kx$ cuts the graph of $y = \frac{2x^2+3}{x-2}$ at most twice, the maximum number of real roots is two.

ACJC Tutorial 2 Q8

ACJC Tutorial 2 Q8

Video Solution: