Chapter 1.2, Problem 78AYU

In Problems 73-80, (a) find the intercepts of each equation, (b) test each equation for symmetry with respect to the x-axis, the y-axis, and the origin, and (c) graph each equation by hand by plotting points. Be sure to label the intercepts on the graph and use any symmetry to assist in drawing the graph. Verify your results using a graphing utility.

$x^2+y^2=16$

To determine

To find: The intercepts of the equation $x^2+y^2=16,$ symmetry with respect to $x\text{-axis}$, $y\text{-axis}$ and origin and graph of the equation.

Answer to Problem 78AYU

$X\text{-intercepts}A\left(-4,0\right),B\left(4,0\right)$

$Y\text{-intercept}\text{is}C\left(0,4\right),D\left(0,-4\right)$

Explanation of Solution

Given:

The equation $x^2+y^2=16$

Calculation:

To find $x\text{-intercept}$, let $y=0$ in the equation

$x^2+{(0)}^2=16$

$x=\pm 4$

$X\text{-intercept}$ are $\left(4,0\right)$ and $\left(-4,0\right)$

To find $x\text{-intercept}$, let $x=0$ in the equation

${(0)}^2+y^2=16$

$y^2=16$

$y=\pm 4$

$Y\text{-intercepts}$ is $\left(0,-4\right)$ and $\left(0,4\right)$

To find the symmetry with respect to $x\text{-axis}$, replace $y$ by $-y$ in the equation

$x^2+{(-y)}^2=16$

$x^2+y^2=16$ which is equivalent to the original equation. The graph of the equation is symmetric with respect to the $x\text{-axis}$.

To find the symmetry with respect to $y\text{-axis}$, replace $x$ by $-x$ in the equation

${(-x)}^2+y^2=16$

$x^2+y^2=16$ which is equivalent to the original equation. The graph of the equation is symmetric with respect to the $y\text{-axis}$.

To find the symmetry with respect to the origin, replace $x$ by $-x$ and $y$ by $-y$ in the equation

${(-x)}^2+{(-y)}^2=16$

$x^2+y^2=16$ which is equivalent to the original equation. The graph of the equation is symmetric with respect to the origin.