Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 5 Polynomials and Polynomial Functions - 5.7 Apply the Fundamental Theorem of Algebra - 5.7 Exercises - Mixed Review - Page 386: 78



Work Step by Step

Here, we find the x-intercepts are $2$ and $7$. Thus, the function is: $y=a (x-2)(x-7)$ Since $(4,-2)$ is on the graph of this function, we have $-2=a (x-2)(x-7)$ This gives: $-2=-6a \implies a=\dfrac{1}{3}$ Hence, the required function is: $y=\dfrac{1}{3}(x-2)(x-7)$
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.