<ol class="i-list"><li><div class="list-item-paragraph">Sketch the curve with equation $y=\left| \frac{\alpha x}{x+1} \right|$, where&nbsp;$\alpha $ is a positive constant, stating the equations of the asymptotes. On the same diagram, sketch the line with equation $y=\alpha x-2$.<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">Solve the inequality $\left| \frac{\alpha x}{x+1} \right|\ge \alpha x-2$, giving your answers in term of $\alpha $.<span class="inline-marks">[3]</span></div></li></ol>

Sketch the curve with equation $y=\left| \frac{\alpha x}{x+1} \right|$, where $\alpha $ is a positive constant, stating the equations of the asymptotes. On the same diagram, sketch the line with equation $y=\alpha x-2$.

Solve the inequality $\left| \frac{\alpha x}{x+1} \right|\ge \alpha x-2$, giving your answers in term of $\alpha $.

Parametric Equations

Graphing Techniques

H2 Math

International Baccalaureate Higher Level Mathematics

IB HL Math

H1 Math

2019 TJC Promo Q1

For \$y = 0\$: \$ax - 2 = 0 \\Rightarrow x = \\dfrac{2}{a}\$

The graph of \$y = \\left|\\dfrac{ax}{x+1}\\right|\$ is the graph of \$y = \\dfrac{ax}{x+1}\$ with the portion below the x-axis reflected above.

Key points: Origin \$(0, 0)\$, horizontal asymptote \$y = a\$, vertical asymptote \$x = -1\$

Line \$y = ax - 2\$ passes through \$(0, -2)\$ and \$\\left(\\dfrac{2}{a}, 0\\right)\$

\$x = \\dfrac{-(-2) \\pm \\sqrt{(-2)^2 - 4(a)(-2)}}{2a} = \\dfrac{2 \\pm \\sqrt{4 + 8a}}{2a} = \\dfrac{2 \\pm \\sqrt{4(1 + 2a)}}{2a}\$

\$= \\dfrac{2 \\pm 2\\sqrt{1 + 2a}}{2a} = \\dfrac{1 \\pm \\sqrt{1 + 2a}}{a}\$

\$x = \\dfrac{1 + \\sqrt{1 + 2a}}{a}\$ or \$x = \\dfrac{1 - \\sqrt{1 + 2a}}{a}\$

Since \$x > 0\$, reject \$x = \\dfrac{1 - \\sqrt{1 + 2a}}{a}\$

Intersection at \$x = \\dfrac{1 + \\sqrt{1 + 2a}}{a}\$

\$\\therefore x \\leq \\dfrac{1 + \\sqrt{1 + 2a}}{a}\$ and \$x \\neq -1\$

2019 TJC Promo Q1

2019 TJC Promo Q1

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