# How do you write the equation of the parabola in vertex form given vertex (4,-1) and y intercept (0,15)?

Aug 30, 2017

See below.

#### Explanation:

Equation in vertex form is $a {\left(x - h\right)}^{2} + k$ where $h$ is the axis of symmetry and $k$ is maximum or minimum value.
In this example $h$ is $- 4$ and $k$ is $- 1$:
So
$a {\left(x + 4\right)}^{2} - 1$ writing as a function: $y = a {\left(x + 4\right)}^{2} - 1$

$y$ intercept is $\left(0 , 15\right)$:

Hence: $y = a {\left(x + 4\right)}^{2} - 1$ => $15 = a {\left(0 + 4\right)}^{2} - 1$
Solving for $a$ gives $a = \frac{15}{16}$

$\frac{15}{16} {\left(x + 4\right)}^{2} - 1$

Aug 30, 2017

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

#### Explanation:

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

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

$\text{here } \left(h , k\right) = \left(4 , - 1\right)$

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

$\text{to find a substitute "(0,15)" into the equation}$

$15 = 16 a - 1 \Rightarrow a = 1$

$\Rightarrow y = {\left(x - 4\right)}^{2} - 1 \leftarrow \textcolor{red}{\text{ in vertex form}}$
graph{(x-4)^2-1 [-10, 10, -5, 5]}