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: 23

Answer

a. $-4(x+\frac{3}{2})^{2} + 10$ b. $Vertex: (\frac{-3}{2},10)$ $x-intercepts: x= \frac{-3-\sqrt 10}{2}$ and $x= \frac{-3+\sqrt 10}{2}$ $y-intercept: f(0) = -4\times0^{2} - 12\times0 + 1 = 1$ c. See the graph d. Domain is all real numbers and Range is from (-infinity, 10]
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Work Step by Step

a. $-4x^{2} - 12x + 1 = -4(x^{2} + 3x + \frac{9}{4}) – \frac{9}{4}\times(-4) + 1 = -4(x+\frac{3}{2})^{2} + 10$ b. $Vertex: (\frac{-3}{2},10)$ $x-intercepts: f(x)=0$ $-4x^{2} - 12x + 1 = 0$ => $x= \frac{-(-12) +\sqrt ((-12)^{2}-4\times(-4)\times1)}{2\times(-4)}$ and $x= \frac{-(-12) -\sqrt ((-12)^{2}-4\times(-4)\times1)}{2\times(-4)}$ => $x= \frac{12 +\sqrt (144+16)}{-8}$ and $x= \frac{12 -\sqrt (144+16)}{-8}$ => $x= \frac{-3-\sqrt 10}{2}$ and $x= \frac{-3+\sqrt 10}{2}$ $y-intercept: f(0) = -4\times0^{2} - 12\times0 + 1 = 1$ c. See the graph d. From the graph Domain is all real numbers Range is from (-infinity, 10]
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