# How do you find the vertex and intercepts for y=1/2(x-8)^2+2?

Mar 22, 2018

$V e r t e x = \left(8 , 2\right)$
$y \text{-intercept:} \left(0 , 34\right)$
$x \text{-intercept: None}$

#### Explanation:

Quadratic equations are either shown as:

$f \left(x\right) = a {x}^{2} + b x + c$ $\textcolor{b l u e}{\text{ Standard Form}}$

$f \left(x\right) = a {\left(x - h\right)}^{2} + k$ $\textcolor{b l u e}{\text{ Vertex Form}}$

In this case, we'll ignore the $\text{standard form}$ due to our equation being in $\text{vertex form}$

$\text{Vertex form}$ of quadratics is much easier to graph due to there not being a need to solve for the vertex, it's given to us.

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

$\frac{1}{2} = \text{Horizontal stretch}$
$8 = x \text{-coordinate of vertex}$
$2 = y \text{-coordinate of vertex}$

It's important to remember that the vertex in the equation is $\left(- h , k\right)$ so since h is negative by default, our $- 8$ in the equation actually becomes positive. That being said:

Vertex = color(red)((8, 2)

Intercepts are also very easy to calculate:

$y \text{-intercept:}$

$y = \frac{1}{2} {\left(0 - 8\right)}^{2} + 2$ $\textcolor{b l u e}{\text{ Set " x=0 " in the equation and solve}}$

$y = \frac{1}{2} {\left(- 8\right)}^{2} + 2$ $\textcolor{b l u e}{\text{ } 0 - 8 = - 8}$

$y = \frac{1}{2} \left(64\right) + 2$ $\textcolor{b l u e}{\text{ } {\left(- 8\right)}^{2} = 64}$

$y = 32 + 2$ $\textcolor{b l u e}{\text{ } \frac{1}{2} \cdot \frac{64}{1} = \frac{64}{2} = 32}$

$y = 34$ $\textcolor{b l u e}{\text{ } 32 + 2 = 4}$

$y \text{-intercept:}$ color(red)((0, 34)

$x \text{-intercept:}$

$0 = \frac{1}{2} {\left(x - 8\right)}^{2} + 2$ $\textcolor{b l u e}{\text{ Set " y=0 " in the equation and solve}}$

$- 2 = \frac{1}{2} {\left(x - 8\right)}^{2}$ $\textcolor{b l u e}{\text{ Subtract 2 from both sides}}$

$- 4 = {\left(x - 8\right)}^{2}$ $\textcolor{b l u e}{\text{ Divide both sides by } \frac{1}{2}}$

$\sqrt{- 4} = \sqrt{{\left(x - 8\right)}^{2}}$ $\textcolor{b l u e}{\text{ Square-rooting both removes the square}}$

$x \text{-intercept:}$ $\textcolor{red}{\text{No Solution}}$ $\textcolor{b l u e}{\text{ Can't square root negative numbers}}$

You can see this to be true, as there are no $x \text{-intercepts:}$ )