Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 4 Quadratic Functions and Factoring - 4.10 Write Quadratic Functions and Models - 4.10 Exercises - Skill Practice - Page 313: 38


See below

Work Step by Step

The standard form of the equation is: $y=ax^2+bx+c$ Given three points: $(-6,29)\\(-4,12)\\(2,-3)$ Substitute: $29=a(-6)^2+b(-6)+c\\12=a(-4)^2+b(-4)+c\\-3=a(2)^2+b(2)+c$ We have the system: $36a-6b+c=29\\16a-4b+c=12\\4a+2b+c=-3$ Subtract the third equation from the second equation: $12a-6b=15\\ \rightarrow 4a-2b=5$ (1) Subtract the third equation from the first equation: $32a-8b=32\\ \rightarrow 4a-b=4$ (2) Subtract equation (1) from equation (2): $b=-1$ Substitute $b$: $4a-(-1)=4\\a=\frac{3}{4}$ Find $c$: $4(\frac{3}{4})+2(-1)+c=-3\\ \rightarrow c=-4$ Hence, $a=\frac{3}{4}\\b=-1\\c=-4$ Substitute back to the initial equation: $y=\frac{3}{4}x^2-x-4$
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.