Chapter 6.1, Problem 69E

To determine

To find: whether graph of equation is symmetric with respect to x- axis, y- axis, the line $y=x$ , the line $y=-x$ , or none of these.

Answer to Problem 69E

Symmetric to y- axis, line $y=x$ , and line $y=-x$

Explanation of Solution

Given information:

  $\begin{array}{c}{x}^2+y^2=16\\ y=\sqrt{16-x^2}\left[\text{Rewritten}\right]\end{array}$

Calculation:

Put $\left(a,b\right)$ into the equation and solve as follows:

  $\begin{array}{l}y=\sqrt{16-x^2}\\ b=\sqrt{16-a^2}\end{array}$

If the solution of equation is same as that of any other case, then their graphs are symmetric.

Let’s check whether the given cases are symmetric with the given equation or not by substituting replacing $\left(x,y\right)$with different coordinates in the equation and simplify as shown below.

Check symmetry with	$\left(x,y\right)$	$y=\sqrt{16-x^2}$	Symmetric of graph equation (Yes/No)
x- axis	$\left(a,-b\right)$	$\begin{array}{c}-b=\sqrt{16-a^2}\\ b=-\sqrt{16-a^2}\end{array}$	No
y- axis	$\left(-a,b\right)$	$\begin{array}{c}b=\sqrt{16-{\left(-a\right)}^2}\\ b=\sqrt{16-a^2}\end{array}$	Yes
$y=x$	$\left(b,a\right)$	$\begin{array}{l}a=\sqrt{16-b^2}\\ a^2=16-b^2\\ b^2=16-a^2\\ b=\sqrt{16-a^2}\end{array}$	Yes
$y=-x$	$\left(-b,-a\right)$	$\begin{array}{l}-a=\sqrt{16-{\left(-b\right)}^2}\\ -a=\sqrt{16-b^2}\\ a^2=16-b^2\\ b^2=16-a^2\\ b=\sqrt{16-a^2}\end{array}$	Yes

Thus the graph of equation is symmetric with y- axis, line $y=x$ , and line $y=-x$ .