Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter F - Foundations: A Prelude to Functions - Section F.2 Graphs of Equations in Two Variables; Intercepts; Symmetry - F.2 Assess Your Understanding - Page 18: 76


$a=-1$ or $a=-5$

Work Step by Step

If a point is on the graph, the $(x,y)$ coordinates satisfy the equation. Here, $(a,-5)$ is on the graph, therefore we should plug in these values into the given equation to obtain: $y=x^2+6x$ $-5=a^2+6a$ $0=a^2+6a+5$ By using the quadratic formula: $a_{1,2}=\dfrac{-6\pm\sqrt{6^2-4(1)(5)}}{2(1)}=\dfrac{-6\pm\sqrt{36-20}}{2}=\dfrac{-6\pm4}{2}$ $a_1=\dfrac{-6+4}{2}=-1$ $a_2=\dfrac{-6-4}{2}=-5$ The points $(-1,-5)$ and $(-5,-5)$ are on the graph.
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.