Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 2 - Graphs and Functions - 2.1 Rectangular Coordinates and Graphs - 2.1 Exercises - Page 194: 52

Answer

$\bf \color {blue}{(0,2),(1,3),(-1,3)}$ $\bf \color {blue}{\text{or any other point on the parabola}}$

Work Step by Step

$$\bf{y=x^2+2}$$ $\bf \text{Lets plug in numbers to find three points}$ if $x=0$: $y=(0)^2+2$ $y=2$ $\bf \color {blue}{(0,2)}$ if $x=1$: $y=1^2+2$ $y=1+2$ $y=3$ $\bf \color {blue}{(1,3)}$ if $x=-1$: $y=(-1)^2+2$ $y=1+2$ $y=3$ $\bf \color {blue}{(-1,3)}$ So three of the points are: $\bf \color {blue}{(0,2),(1,3),(-1,3)}$ Which we can graph as a parabola
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