Rational Function with a Parameter: Asymptotes, Extrema, Intercepts

Grades 11-12

Question: y=\frac{ax^{2}+x-3}{x^{3}-6}. give a slide of a. draw the graph. locate points of inflexion, extrema, asymptotes, xy intercepts on the graph

To analyze the graph of the rational function

y = \frac{ax^{2}+x-3}{x^{3}-6}

, we must apply techniques from the HKDSE Mathematics Extended Part (Module 1/Module 2). This specific problem is considered out-of-syllabus for the Compulsory Part as it requires the study of rational functions where the degree of the denominator is higher than the numerator, as well as the identification of inflection points via the second derivative.

1. Asymptotes (Calculus of Limits)

Vertical Asymptote (V.A.): These occur where the denominator is zero and the numerator is non-zero. $\begin{align*} x^3 - 6 &= 0 \\ x &= \sqrt[3]{6} \approx 1.817 \end{align*}$ As $x \to \sqrt[3]{6}$ , $|y| \to \infty$ . [Reason: infinite limit]
Horizontal Asymptote (H.A.): We examine the limit as $x \to \pm\infty$ . $\begin{align*} \lim_{x \to \infty} \frac{ax^2+x-3}{x^3-6} &= \lim_{x \to \infty} \frac{\frac{a}{x} + \frac{1}{x^2} - \frac{3}{x^3}}{1 - \frac{6}{x^3}} = \frac{0}{1} = 0 \end{align*}$ The horizontal asymptote is the line $y = 0$ (the $x$ -axis).

2. Intercepts

$y$ -intercept: Set $x=0$ . $\begin{align*} y &= \frac{a(0)^2 + 0 - 3}{0^3 - 6} = \frac{-3}{-6} = 0.5 \end{align*}$ The graph always passes through $(0, 0.5)$ regardless of the value of $a$ .
$x$ -intercepts: Set y=0, which implies the numerator ax^2 + x - 3 = 0. Using the quadratic formula: x = \frac{-1 \pm \sqrt{1 + 12a}}{2a}.
- If $a > -\frac{1}{12}$ , there are two $x$ -intercepts.
- If $a = -\frac{1}{12}$ , there is one $x$ -intercept.
- If $a < -\frac{1}{12}$ , there are no real $x$ -intercepts.

3. Extrema and Inflection Points

To locate these exactly, we differentiate using the Quotient Rule:

y' = \frac{(x^3-6)(2ax+1) - (ax^2+x-3)(3x^2)}{(x^3-6)^2}

Extrema (Stationary Points): Occur when $y' = 0$ . This leads to a quartic equation in $x$ . Visually, these are the "peaks" and "valleys" on the curve.
Points of Inflection: Occur where $y'' = 0$ and the concavity changes. These are the points where the curve changes from "cupping up" to "cupping down".

In the provided interactive diagram, you can adjust the slider a to see how the curvature and the

x

-intercepts change. Note how the vertical asymptote at

x \approx 1.82

remains fixed, acting as a boundary for the function's branches.