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 591: 40

Answer

$x^2=-y$

Work Step by Step

Since the parabola is symmetric with respect to the $y-$axis, it means that the parabola is opening up or down. The standard form of a vertical parabola is given as: $x^2=4py ~~~(1)$ (when the parabola has the vertex at origin $(0,0)$) Here, the point $(2, -4)$ is on the graph, so we have: $x=2; y= -4$ Now, we will plug the values of $x$ and $y$ in the standard form of parabola to determine the value of $p$: $(2)^2=4p \times (-4)$ or, $-16p=4 \implies p=-\frac{1}{4}$ Thus, the equation (1) becomes: $x^2 =(4)\left(-\frac{1}{4}\right) y \implies x^2=-y$
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