# What is the equation of the parabola that has a vertex at  (5, 4)  and passes through point  (7,-8) ?

Oct 9, 2017

The equation of parabola is $y = - 3 {x}^{2} + 30 x - 71$

#### Explanation:

The equation of parabola in vertex form is $y = a {\left(x - h\right)}^{2} + k$

$\left(h , k\right)$ being vertex here $h = 5 , k = 4 \therefore$ Equation of parabola in

vertex form is $y = a {\left(x - 5\right)}^{2} + 4$ . The parabola passes through

point $\left(7 , - 8\right)$ . So the point $\left(7 , - 8\right)$ will satisfy the equation .

$\therefore - 8 = a {\left(7 - 5\right)}^{2} + 4 \mathmr{and} - 8 = 4 a + 4$ or

$4 a = - 8 - 4 \mathmr{and} a = - \frac{12}{4} = - 3$ Hence the equation of

parabola is $y = - 3 {\left(x - 5\right)}^{2} + 4$ or

$y = - 3 \left({x}^{2} - 10 x + 25\right) + 4 \mathmr{and} y = - 3 {x}^{2} + 30 x - 75 + 4$ or

$y = - 3 {x}^{2} + 30 x - 71$

graph{-3x^2+30x-71 [-20, 20, -10, 10]}

Oct 9, 2017

$y = - 3 {x}^{2} + 30 x - 71$

#### Explanation:

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

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

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

$\text{here } \left(h , k\right) = \left(5 , 4\right)$

$\Rightarrow y = a {\left(x - 5\right)}^{2} + 4$

$\text{to find a substitute "(7,-8)" into the equation}$

$- 8 = 4 a + 4 \Rightarrow a = - 3$

$\Rightarrow y = - 3 {\left(x - 5\right)}^{2} + 4 \leftarrow \textcolor{red}{\text{ in vertex form}}$

$\text{distributing and simplifying gives}$

$y = - 3 \left({x}^{2} - 10 x + 25\right) + 4$

$\textcolor{w h i t e}{y} = - 3 {x}^{2} + 30 x - 75 + 4$

$\Rightarrow y = - 3 {x}^{2} + 30 x - 71 \leftarrow \textcolor{red}{\text{ in standard form}}$