# How do you write the vertex form equation of each parabola given Vertex (8, -1), y-intercept: -17?

Mar 23, 2018

Equation of parabola is y = -1/4(x-8)^2-1 ;

#### Explanation:

Vertex form of equation of parabola is y = a(x-h)^2+k ; (h,k)

being vertex. Here $h = 8 , k = - 1$ Hence equation of parabola is

y = a(x-8)^2-1 ; , y intercept is $- 17 \therefore \left(0 , - 17\right)$ is a point

through which parabola passes , so the point will satisfy the

equation of parabola $- 17 = a {\left(0 - 8\right)}^{2} - 1 \mathmr{and} - 17 = 64 a - 1$

or $64 a = - 16 \mathmr{and} a = - \frac{16}{64} = - \frac{1}{4}$. So equation of parabola

is y = -1/4(x-8)^2-1 ;

graph{-1/4(x-8)^2-1 [-40, 40, -20, 20]} [Ans]

Mar 24, 2018

$y = - \frac{1}{4} {\left(x - 8\right)}^{2} - 1$

#### Explanation:

$\text{the equation of a parabola in "color(blue)"vertex form}$

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

$\text{where "(h,k)" are the coordinates of the vertex and a}$
$\text{is a multiplier}$

$\text{here } \left(h , k\right) = \left(8 , - 1\right)$

$\Rightarrow y = a {\left(x - 8\right)}^{2} - 1$

$\text{to find a substitute "(0,-17)" into the equation}$

$- 17 = 64 a - 1$

$\Rightarrow 64 a = - 16 \Rightarrow a = - \frac{1}{4}$

$\Rightarrow y = - \frac{1}{4} {\left(x - 8\right)}^{2} - 1 \leftarrow \textcolor{red}{\text{in vertex form}}$