# What is the equation of the parabola that has a vertex at  (0, 8)  and passes through point  (5,-4) ?

Jan 22, 2016

There are an infinite number of parabolic equations that meet the given requirements.
If we restrict the parabola to having a vertical axis of symmetry, then:
$\textcolor{w h i t e}{\text{XXX}} y = - \frac{12}{25} {x}^{2} + 8$

#### Explanation:

For a parabola with a vertical axis of symmetry, the general form of the parabolic equation with vertex at $\left(a , b\right)$ is:
$\textcolor{w h i t e}{\text{XXX}} y = m {\left(x - a\right)}^{2} + b$

Substituting the given vertex values $\left(0 , 8\right)$ for $\left(a , b\right)$ gives
$\textcolor{w h i t e}{\text{XXX}} y = m {\left(x - 0\right)}^{2} + 8$

and if $\left(5 , - 4\right)$ is a solution to this equation, then
$\textcolor{w h i t e}{\text{XXX}} - 4 = m \left({\left(- 5\right)}^{2} - 0\right) + 8 \Rightarrow m = - \frac{12}{25}$

and the parabolic equation is
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l a c k}{y = - \frac{12}{25} {x}^{2} + 8}$
graph{y=-12/25*x^2+8 [-14.21, 14.26, -5.61, 8.63]}

However, (for example) with a horizontal axis of symmetry:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l a c k}{x = \frac{5}{144} {\left(y - 8\right)}^{2}}$
also satisfies the given conditions:
graph{x=5/144(y-8)^2 [-17.96, 39.76, -8.1, 20.78]}

Any other choice for the slope of the axis of symmetry will give you another equation.