Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 9 - Section 9.3 - The Parabola - Exercise Set - Page 997: 34

Answer

$\text{Vertex}= (1,1)$ Focus: $f(0,1)$ Directrix is: $x=2$ 'Graph b'

Work Step by Step

General form of a horizontal parabola is given as: $(y-k)^2=4p(x-h)$...(1) Here, $\text{Vertex}=(h,k)$ and focus is: $(h+p, k)$ As we are given $(y-1)^2=-4(x-1)$ From equation (1), we get: $k=1, h=1$ and $4p=-4 \implies p=-1$ $\text{Vertex}=(h,k) \implies (1,1)$ Also, focus is: $(h+p,k) \implies f(1-1,1)=f(0,1)$ Directrix is: $x=h-p \implies x=1+1=2$ Hence, $\text{Vertex}= (1,1)$ Focus: $f(0,1)$ Directrix is: $x=2$ Thus, the 'Graph b' matches with all the above conditions.
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.