## Algebra and Trigonometry 10th Edition

$x^2 =-\dfrac{25}{4} (y-48)$
We need to write the Standard form of a parabola with a vertical axis as follows: $(x-h)^2 =4p (y-k)$ Now, $(x-0)^2 =4p (y-48)$ $h=0$ and $k=58$ Vertex: $(0, 48)$ Points are: $(10 \sqrt 3,0)$ $(10\sqrt 3-0)^2 =4p (0-48) \implies p=-\dfrac{25}{16}$ Back substitution: $(x-0)^2 =4(-\dfrac{25}{16}) (y-48)$ Our answer is: $x^2 =-\dfrac{25}{4} (y-48)$