College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 7 - Section 7.2 - The Parabola - 7.2 Assess Your Understanding - Page 515: 44

Answer

vertex: $\quad (-4,-2)$ focus: $\quad (-4,2)$ directrix: $\quad y=-6$

Work Step by Step

Comparing the form of the equation with table 2, $\quad \quad \quad (x+4)^{2}=16(y+2)\quad $has the form $\begin{array}{|c|c|c|c|} \hline {focus}&{directrix}&{equation}& ... opens\\ \hline {(h,k+a)}&{y=k-a}&{(x-h)^{2}=4a(y-k)}&up\\ \hline \end{array}$ $(x+4)^{2}=4(4)(y+2)\quad \Rightarrow\quad h=-4, k=-2, a=4$ vertex: $\quad (h,k)=(-4,-2)$ focus: $\quad (h,k+a)=(-4,-2+4)=(-4,2)$ directrix: $\quad y=k-a\quad \Rightarrow\quad y=-6$ To graph, find the endpoints of the latus rectum (line segment parallel to the directrix, containing the focus) by setting $y=2$ $(x+4)^{2}=16(2+2)$ $(x+4)^{2}=64$ $x+4=\pm 8$ $x=-4\pm 8$ Endpoints of the latus rectum: $(-12,2), (4,2)$
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