College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 6 - Section 6.1 - Parabolas - 6.1 Exercises - Page 590: 38

Answer

$y^2=-4x$

Work Step by Step

Since, the parabola is opening left, it means that the parabola is a horizontal. The standard form of a horizontal parabola is given as: $y^2=4px ~~~(1)$ (when the parabola has the vertex at origin $(0,0)$) Here, the point $(-2, -2\sqrt 2)$ is on the graph, so we have: $x=-2; y=-2 \sqrt 2$ Now, we will plug the values of $x$ and $y$ in the standard form of parabola to determine the value of $p$: $(-2\sqrt 2)^2=4p \times (-2)$ or, $-8p=8 \implies p=-1$ Thus, the equation (1) becomes: $y^2 =(4)(-1) x \implies y^2=-4x$
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