Chapter 2, Problem 14CT

To determine

Find intercepts and asymptotes of the given function also sketch the graph.

Answer to Problem 14CT

  $x$ -intercept of $h(x)=\frac{3}{x^2}-1$ is $[(-\sqrt{3},0),(\sqrt{3},0)]$

  $y$ -intercept of $h(x)=\frac{3}{x^2}-1$ approaches infinity

  $x=0$ is a vertical asymptote.

  $y=-1$ is horizontal asymptote.

Explanation of Solution

Given:

logarithmic function:

  $h(x)=\frac{3}{x^2}-1$

To graph the given logarithmic function

  $x$ -intercept of $h(x)=\frac{3}{x^2}-1$ is obtained by putting $h(x)=0$ in $h(x)=\frac{3}{x^2}-1$ and solving for $x$

  $0=\frac{3}{x^2}-1$

  $\Rightarrow 1=\frac{3}{x^2}$ Add 1 to both sides

  $\Rightarrow x^2=3$ Multiply both sides by $x^2$

  $\Rightarrow x=\pm \sqrt{3}$ Square root property

Therefore, $x$ -intercept of $h(x)=\frac{3}{x^2}-1$ is $[(-\sqrt{3},0),(\sqrt{3},0)]$

  $y$ -intercept of $h(x)=\frac{3}{x^2}-1$ is obtained by putting $x=0$ in $h(x)=\frac{3}{x^2}-1$ and solving for $h(x)$

But at $x=0,\frac{3}{x^2}$ approaches infinity and hence $h(x)=\frac{3}{x^2}-1$ approaches infinity.

Therefore, $x=0$ is a vertical asymptote.

Also, as $x$ tends to infinity, $\frac{3}{x^2}$ approaches zero and hence $h(x)=\frac{3}{x^2}-1$ approaches -1.

Therefore, $y=-1$ is horizontal asymptote.

Graph of $h(x)=\frac{3}{x^2}-1$ is shown below;