Chapter 1.8, Problem 71E

To determine

To calculate: The x- and y- intercepts for the equation $x+y^2=4$ and test the symmetry of the equation. Also construct the table to sketch the graph of the equation.

Answer to Problem 71E

The x-intercept is $4$ and y-intercepts are $\pm 2$ . The equation is symmetric about x-axis.Graph of the equation $x+y^2=4$ is provided below,

  

Explanation of Solution

Given information:

The equation $x+y^2=4$ .

Formula used:

The function is symmetric about the x-axis, when y is replaced by $-y$ , the equation remains unchanged.

The function is symmetric about the y-axis, when x is replaced by $-x$ , the equation remains unchanged.

The function is symmetric with respect to origin, when y is replaced by $-y$ and x is replaced by $-x$ , the equation remains unchanged.

The x-intercepts are the points on x-axis where the graph of the equation intersects the x-axis.

The y-intercepts are the points on y-axis where the graph of the equation intersects the y-axis.

Calculation:

It is provided that the equation is $x+y^2=4$ . Construct a table to evaluate the value of y for different values of x.

Substitute the point $x=0$ in the equation $x+y^2=4$ ,

  $\begin{array}{c}0+y^2=4\\ y^2=4\\ y=\pm 2\end{array}$

Substitute the point $x=3$ in the equation $x+y^2=4$ ,

  $\begin{array}{c}3+y^2=4\\ y^2=1\\ y=\pm 1\end{array}$

Construct a table with the values obtained above,

  $\begin{array}{ccc}x& y& \left(x,y\right)\\ 0& 2& \left(0,-2\right)\\ 0& -2& \left(0,2\right)\\ 3& -1& \left(3,-1\right)\\ 3& 1& \left(3,1\right)\end{array}$

In the coordinate plane plot the points obtained above and connect them through a line.

The graph of the equation is provided below $x+y^2=4$ .

  

Recall that the x-intercepts are the points on x-axis where the graph of the equation intersects the x-axis.

Substitute $y=0$ in the equation $x+y^2=4$ ,

  $\begin{array}{c}x+0^2=4\\ x=4\end{array}$

Therefore, x-intercept is 4.

Recall that the y-intercepts are the points on x-axis where the graph of the equation intersects the y-axis.

Substitute $x=0$ in the equation $x+y^2=4$ ,

  $\begin{array}{c}0+y^2=4\\ y^2=4\\ y=\pm 2\end{array}$

Therefore, y-intercept are $\pm 2$ .

Recall that the function is symmetric about the x-axis, when y is replaced by $-y$ , the equation remains unchanged.

Replace y by $-y$ in the equation $x+y^2=4$ ,

  $\begin{array}{c}x+{\left(-y\right)}^2=4\\ x+y^2=4\end{array}$

The equation is unchanged. Therefore, the equation $x+y^2=4$ is symmetricabout the x-axis.

Recall that the function is symmetric about the y-axis, when x is replaced by $-x$ , the equation remains unchanged.

Replace x by $-x$ in the equation $x+y^2=4$ ,

  $-x+y^2=4$

The equation is changed. Therefore, the equation $x+y^2=4$ is not symmetricabout the y-axis.

Recall that the function is symmetric with respect to origin, when y is replaced by $-y$ and x is replaced by $-x$ , the equation remains unchanged.

Replace x by $-x$ and y by $-y$ in the equation $x+y^2=4$ ,

  $\begin{array}{c}-x+{\left(-y\right)}^2=4\\ -x+y^2=4\end{array}$

The equation is changed. Therefore, the equation $x+y^2=4$ is not symmetricabout the origin.

Thus, the x-intercept is 4 and y-interceptsare $\pm 2$ . The equation is symmetric about y-axis.Graph of the equation $x+y^2=4$ is provided below,