# What is the vertex of the parabola y=1/8(x-2)^2+5?

May 2, 2016

$\left(2 , 5\right)$

#### Explanation:

The equation:

$y = \frac{1}{8} {\left(x - 2\right)}^{2} + 5$

is in vertex form:

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

with $a = \frac{1}{8}$ and $\left(h , k\right) = \left(2 , 5\right)$

So we simply read the coordinates of the vertex $\left(h , k\right) = \left(2 , 5\right)$ from the coefficients of the equation.

Notice that for any Real value of $x$, the resulting value of ${\left(x - 2\right)}^{2}$ is non-negative, and it is only zero when $x = 2$. So this is where the vertex of the parabola is.

When $x = 2$, the resulting value of $y$ is ${0}^{2} + 5 = 5$.

graph{(1/8(x-2)^2+5-y)((x-2)^2+(y-5)^2-0.03)=0 [-14.05, 17.55, -1.89, 13.91]}