Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 3 - Linear and Quadratic Functions - Chapter Review - Chapter Test - Page 164: 6


(a) opens up (b) $(2,-8)$ (c) $x=2$ (d) The $y$-intercept is $4$; the $x$-intercepts are $\dfrac{6-2\sqrt6}{3}$ and $\dfrac{6+2\sqrt3}{6}$; (e) Refer to the graph below.

Work Step by Step

a) Identify $a,b,c$ from the standard form $f(x)=ax^2+bx+c$: $a=3$ $b=-12$ $c=4$ Since $a$ is positive, then the parabola opens up. b) Determine the vertex using the formula $\left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)$ $x=-\dfrac{b}{2a}=-\dfrac{-12}{2(3)}=2$ $y=3(2^2)-12(2)+4=-8$ Therefore the vertedx is at $(2,-8)$. c) The axis of symmetry is the line $x=-\frac{b}{2a}$.: Thus, the axis of symmetry is $x=2$. d) Determine the intercepts: FInd the $y$-intercept by setting $x=0$ then solving for the value of $y$: $y=3(0^2)-12(0)+4=4$ Find the $x$-intercept/s by setting $y=0$ then solving the value of $x$ using the quadratic formula: $3x^2-12x+4=0$ (so $a=3, b=-12, $ and $c=4$) $x=\dfrac{12\pm\sqrt{(-12)^2-4(3)(4)}}{2(3)}\\ x=\dfrac{12\pm\sqrt{144-48}}{6}\\ x=\dfrac{12\pm\sqrt{96}}{6}\\ x=\dfrac{12\pm\sqrt{16(6)}}{6}\\ x=\dfrac{12\pm 4\sqrt 6}{6}\\ x=\dfrac{6\pm 2\sqrt 6}{3}$ $x_1=\dfrac{6-2\sqrt 6}{3}$ $x_2=\dfrac{6+2\sqrt 6}{3}$ e) Graph the function: Plot the vertex, $x$-intercepts, and the $y$-intercept, then reflect these points about the axis of symmetry $x=2$. Then, connect the points using a smooth curve. Refer to the graph below.
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