Chapter 9.2, Problem 76S

To determine

Graph the function $y={(-x)}^2+1$

Answer to Problem 76S

To draw the graph joins the points $(0,1),(-1,2),(1,2),(-2,5),(2,5)$ with a smooth curve.

Explanation of Solution

Given:

The function $y={(-x)}^2+1$

Concept Used:

The function $y={(-x)}^2+1$

Calculation:

The function $y={(-x)}^2+1$

Find the vertex and the axis of symmetry:

  $\begin{array}{l}y={(}^{-}+1\\ y=x^2+1\\ a=1;b=0\\ x=-\frac{b}{2a}=-\frac{0}{1}=0\\ x=0\end{array}$

  $\begin{array}{l}x=0;y=?\\ y=x^2+1\\ y=0^2+1=1\\ \text{Vertex}=(0,1)\end{array}$ Y − Intercept is $(0,1)$

The axis of symmetry is $x=0$

  $a>0$ , So the graph opens upward and the vertex is a minimum.

Find a couple of points and reflect them across the line of symmetry. $\text{If}x=-1,\text{then}y={(-1)}^2+1=1+1=2$ Plot the point $(-1,2)$ Reflect $(-1,2)$ over the axis of symmetry to obtain the point $(1,2)$ $\text{If}x=-2,\text{then}y={(-2)}^2+1=4+1=5$. Plot the point $(-2,5)$. Reflect $(-2,5)$ over the axis of symmetry to obtain the point $(2,5)$. Connect the points $(0,1),(-1,2),(1,2),(-2,5),(2,5)$ with a smooth curve.

Thus, to draw the graph joins the points $(0,1),(-1,2),(1,2),(-2,5),(2,5)$ with a smooth curve.