Chapter 2, Problem 15CT

To determine

To find: Intercepts and asymptotes of the given function as well as plot graph.

Answer to Problem 15CT

The value of x-intercept is $x=-\frac{3}{2}$ and y-intercept is $y=\frac{3}{4}$

The asymptote is at $y=2$

Explanation of Solution

Given:

The function:

  $\frac{2x^2-5x-12}{x^2-16}=0$

The intercept is when $\frac{2x^2-5x-12}{x^2-16}=0$

First factorize all the factors and see if the factors could be cut or not,

  $\begin{array}{l}\frac{2x^2-5x-12}{x^2-16}=\frac{2x^2-8x+3x-12}{x^2-16}\\ =\frac{2x\left(x-4\right)+3\left(x-4\right)}{x^2-16}\\ =\frac{\left(x-4\right)\left(2x+3\right)}{\left(x-4\right)\left(x+4\right)}\\ =\frac{\left(2x+3\right)}{\left(x+4\right)}\end{array}$

So the value of x-intercept is $x=-\frac{3}{2}$

To find the y intercept let put the value of x equals to 0. So

  $\begin{array}{l}y=\frac{2{\left(0\right)}^2-5\left(0\right)+12}{{\left(0\right)}^2-16}\\ y=\frac{12}{16}\\ y=\frac{3}{4}\end{array}$

The vertical asymptote is when the denominator becomes zero. That is at

  $\begin{array}{l}x+4=0\hfill \\ x=–4\hfill \end{array}$

There will be also a horizontal asymptote. Dividing numerator and denominator by x as follows,

  $\frac{\frac{2x+3}{x}}{\frac{x+4}{x}}=\frac{2+\frac{3}{x}}{1+\frac{4}{x}}$

It will tend to 2 at higher values of x.

So the asymptote is at $y=2$

The graph of the function is as follows,