Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 9 - Quadratic Functions and Equations - Chapter Test - Page 593: 4


Please see the graph.

Work Step by Step

$y=x^2-3x+2$ Axis of symmetry: $x=-b/2a$ $x=-b/2a$ $x=-(-3)/2(1)$ $x=3/2$ $y=x^2-3x+2$ $y=(3/2)^2-3*(3/2)+2$ $y=9/4-9/2+2$ $y=-9/4+2$ $y=-1/4$ $(1.5,-.25)$ $y=x^2-3x+2$ $y=x^2-2x-x+2$ $y=x(x-2)-1(x-2)$ $y=(x-2)(x-1)$ If either $x=2$ or $x=1$, then that part of the equation equals zero, and then the entire formula equals zero. Thus, we have the two points $(1,0)$ and $(2,0)$ on the curve.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.