Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 3 - Section 3.1 - Quadratic Functions and Models - 3.1 Exercises - Page 252: 24

Answer

a. $3(x+\frac{1}{3})^{2} - \frac{7}{3}$ b. $Vertex: (\frac{-1}{3},\frac{-7}{3})$ $x-intercepts: \frac{-1 +\sqrt 7}{3}$ and $x= \frac{-1 -\sqrt 7}{3}$ $y-intercept: -2$ c. See the graph d. Domain: All real numbers Range: $[\frac{-7}{3}, infinity)$
1511852813

Work Step by Step

a. $3x^{2} + 2x - 2 = 3(x^{2} + \frac{2}{3}x + \frac{1}{9}) – \frac{1}{9}\times3 - 2 = 3(x+\frac{1}{3})^{2} - \frac{7}{3}$ b. $Vertex: (\frac{-1}{3},\frac{-7}{3})$ $x-intercepts: f(x)=0$ $3x^{2} + 2x - 2 = 0$ => $x= \frac{-2 +\sqrt (2^{2}-4\times3\times(-2))}{2\times3}$ and $x= \frac{-2 -\sqrt (2^{2}-4\times3\times(-2))}{2\times3}$ => $x= \frac{-2 +\sqrt (4+24)}{6}$ and $x= \frac{-2 -\sqrt (4+24)}{6}$ => $x= \frac{-1 +\sqrt 7}{3}$ and $x= \frac{-1 -\sqrt 7}{3}$ $y-intercept: f(0) = 3\times0^{2} + 2\times0 - 2 = -2$ c. Sketch like in the graph d. Domain is all real numbers Range is from $[\frac{-7}{3}, infinity)$
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.