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: 22

Answer

a. $2(x+3)^{2} - 8$ b. $Vertex: (-3,-8)$ $x-intercepts: -1, -5$ $y-intercept: 10$ c. Sketch like in the graph d. Domain is all real numbers and range is from [-8, infinity)
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Work Step by Step

a. $2x^{2} + 12x + 10 = 2(x^{2} + 6x + 9) - 9\times2 + 10 = 2(x+3)^{2} - 8$ b. $Vertex: (-3,-8)$ $x-intercepts: f(x)=0$ $2x^{2} + 12x + 10 = 0$ => $x= \frac{-12 +\sqrt (12^{2}-4\times2\times10)}{2\times2}$ and $x= \frac{-12 -\sqrt (12^{2}-4\times2\times10)}{2\times2}$ => $x= \frac{-12 +\sqrt (144-80)}{4}$ and $x= \frac{-12 -\sqrt (144-80)}{4}$ => $x= -1$ and $x= -5$ $y-intercept: f(0) = 2\times0^{2} + 12\times0 + 10 = 0$ c. Sketch like in the graph d. Domain is all real numbers Range is from [-8, infinity)
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