# How do you write the quadratic function in vertex form given vertex (4,5) and point (8,-3)?

Oct 3, 2017

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

#### Explanation:

Vertex form looks like this:

$y = a {\left(x - h\right)}^{2} + k$

where $a$ is a constant multiplier affecting the "steepness" of the parabola and whether it is upright or inverted and $\left(h , k\right)$ is the vertex.

So in our example, we are looking for an equation of the form:

$y = a {\left(x - \textcolor{b l u e}{4}\right)}^{2} + \textcolor{b l u e}{5}$

In order that this pass through the point $\left(8 , - 3\right)$ we require that these coordinates satisfy the equation, so:

$\textcolor{b l u e}{- 3} = a {\left(\textcolor{b l u e}{8} - 4\right)}^{2} + 5$

$\textcolor{w h i t e}{- 3} = 16 a + 5$

Subtract $5$ from both ends to get:

$- 8 = 16 a$

Divide both sides by $16$ to find:

$a = - \frac{1}{2}$

So the equation we want is:

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

graph{(y+1/2(x-4)^2-5)((x-4)^2+(y-5)^2-0.01)((x-8)^2+(y+3)^2-0.01) = 0 [-6.21, 13.79, -4.16, 5.84]}