# What is the equation of the parabola that has a vertex at  (-4, 121)  and passes through point  (7,0) ?

$y = - {\left(x + 4\right)}^{2} + 121$

#### Explanation:

Given vertex at $\left(- 4 , 121\right)$ and a point $\left(7 , 0\right)$
$h = - 4$
$k = 121$
$x = 7$
$y = 0$

Use the standard form. Substitute the values to solve for $p$.

${\left(x - h\right)}^{2} = - 4 p \left(y - k\right)$

${\left(7 - - 4\right)}^{2} = - 4 p \left(0 - 121\right)$

${\left(11\right)}^{2} = - 4 p \left(- 121\right)$
$121 = 4 \left(121\right) p$

$\frac{121}{121} = \frac{4 \left(121\right) p}{121}$

$\frac{\cancel{121}}{\cancel{121}} = \frac{4 \left(\cancel{121}\right) p}{\cancel{121}}$

$1 = 4 p$

$p = \frac{1}{4}$

the equation is now

${\left(x - - 4\right)}^{2} = - 4 \left(\frac{1}{4}\right) \left(y - 121\right)$

${\left(x + 4\right)}^{2} = - 1 \left(y - 121\right)$

${\left(x + 4\right)}^{2} = - y + 121$

$y = - {\left(x + 4\right)}^{2} + 121$

graph{y=-(x+4)^2+121[-100,300,-130,130]}

Have a nice day !! from the Philippines.