# How do you write the standard form of the equation of the parabola that has the indicated vertex and whose graph passes through the given point: Vertex: (6, 5); point: (0, -4)?

Jun 18, 2016

$y = - \frac{1}{4} {x}^{2} + 3 x - 4$

#### Explanation:

The $\textcolor{b l u e}{\text{vertex form}}$ of the equation is.

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{y = a {\left(x - h\right)}^{2} + k} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
where (h ,k) are the coordinates of the vertex.

here the vertex = (6 ,5)

$\Rightarrow y = a {\left(x - 6\right)}^{2} + 5 \text{ is the equation}$

To find a , use the point (0 ,-4) that the parabola passes through.

Substitute x = 0 , y = -4 into the equation.

hence : $a {\left(0 - 6\right)}^{2} + 5 = - 4 \Rightarrow 36 a = - 9 \Rightarrow a = - \frac{1}{4}$

$y = - \frac{1}{4} {\left(x - 6\right)}^{2} + 5 \text{ is the vertex form of the parabola}$

The $\textcolor{b l u e}{\text{standard form}}$ of the equation is.

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{y = a {x}^{2} + b x + c} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

To obtain this form expand and simplify the vertex form.

$- \frac{1}{4} {\left(x - 6\right)}^{2} + 5 = - \frac{1}{4} \left({x}^{2} - 12 x + 36\right) + 5$

$= - \frac{1}{4} {x}^{2} + 3 x - 9 + 5 = - \frac{1}{4} {x}^{2} + 3 x - 4$

$\Rightarrow y = - \frac{1}{4} {x}^{2} + 3 x - 4 \text{ is the standard form}$