## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson

# Chapter F - Foundations: A Prelude to Functions - Section F.2 Graphs of Equations in Two Variables; Intercepts; Symmetry - F.2 Assess Your Understanding - Page 16: 22

#### Answer

There are two $x$-intercepts: $-1$ and $1$. The y-intercept is $1$. Refer to the graph below.

#### Work Step by Step

In order to find the $x$-intercept, we have to plug $y=0$ into the equation, and the $x$-coordinate will give us the intercept itself. $y=-x^2+1$ $0=-x^2+1$ $x^2=1$ $x=\pm1$ The same thing should be done for the $y$-intercept, however, here $x=0$ should be plugged into the equation. $y=-x^2+1$ $y=-0^2+1$ $y=1$ There are two $x$-intercepts: $-1$ and $1$. The y-intercept is $1$. Few other points that will help us plot the graph: If $x=2$ $y=-2^2+1=-3$ The point $(2,-3)$ is on the graph. If $x=-2$ $y=-(-2)^2+1=-3$ The point $(-2,-3)$ is on the graph. If $x=3$ $y=-3^2+1=-8$ The point $(3,-8)$ is on the graph.

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