Chapter 1, Problem 116RE

To determine

To graph: The equation $8x+y^2=0$ and check the equation for symmetry.

Answer to Problem 116RE

The equation $8x+y^2=0$ is symmetric about $x$ -axis but not about $y$ -axis and the graph is as follows:

  

Explanation of Solution

Given information:

The equation of line $8x+y^2=0$

Formula used:

The equation is said to have symmetry about the $x$ -axis when the equation remains unchanged on replacing $y$ by $-y$

Graphically it can be said that the graph remains unchanged when reflected in the $x$ -axis.

The equation is said to have symmetry about the $y$ -axis when the equation remains unchanged on replacing $x$ by $-x$

Graphically it can be said that the graph remains unchanged when reflected in the $y$ -axis.

The equation is said to have symmetry about the origin if it is symmetric about $x$ -axis and $y$ -axis that is the equation remains unchanged on replacing $x$ by $-x$ or $y$ by $-y$

Graphically it can be said that the graph remains unchanged when rotated with an angle of ${180}^{\circ }$ about the origin.

Calculation:

It is provided that the equation of line is $8x+y^2=0$ $\left(1\right)$

Firstly replace $y$ by $-y$ in $\left(1\right)$

  $\begin{array}{l}8x+{\left(-y\right)}^2=0\\ 8x+y^2=0\end{array}$

The equation obtained is the same as the original one.

Hence, $8x+y^2=0$ is symmetric about $x$ -axis

Next, replace $x$ by $-x$ in $\left(1\right)$

  $\begin{array}{l}8\left(-x\right)+y^2=0\\ -8x+y^2=0\end{array}$

The equation obtained isnot the same as the original one.

Hence, $8x+y^2=0$ isnot symmetric about $y$ -axis.

For plotting:

$x$	$y=16-x^2$	$\left(x.y\right)$
$\begin{array}{l}0\\ -2\\ -2\end{array}$	$\begin{array}{l}0\\ 4\\ -4\end{array}$	$\begin{array}{l}\left(0,0\right)\\ \left(-2,4\right)\\ \left(-2,-4\right)\end{array}$

We plot these points to get the graph as follows:

  

Thus, $8x+y^2=0$ is symmetric about $x$ -axisbut not about $y$ -axis.