Chapter 1.4, Problem 4QR

To determine

To find: The $x$ and $y$ intercepts of the equation $\frac{x^2}{9}+\frac{y^2}{16}=1$ .

Answer to Problem 4QR

The $x$ intercepts are 3 and $-3$ , and the $y$ -intercepts are 4 and $-4$ .

Explanation of Solution

Given information: The equation is $\frac{x^2}{9}+\frac{y^2}{16}=1$ .

Calculation:

Substitute 0 for $y$ in the given equation to find the $x$ intercepts.

  $\begin{array}{c}\frac{x^2}{9}+\frac{{\left(0\right)}^2}{16}=1\\ \frac{x^2}{9}=1\\ x^2=9\\ x=\pm 3\end{array}$

Substitute 0 for $x$ in the given equation to find the $y$ intercepts.

  $\begin{array}{c}\frac{{\left(0\right)}^2}{9}+\frac{y^2}{16}=1\\ \frac{y^2}{16}=1\\ y^2=16\\ y=\pm 4\end{array}$

Therefore, the $x$ intercepts are 3 and $-3$ , and the $y$ -intercepts are 4 and $-4$ .