# How do you graph y=(x-5)^2?

Jul 17, 2015

This is a vertical parabola - sort of U shaped - with vertex at $\left(5 , 0\right)$, axis $x = 5$ and intercept with the $y$ axis at $\left(0 , 25\right)$.

#### Explanation:

The vertex is at the minimum value of the function.

${\left(x - 5\right)}^{2} \ge 0$ since it is the square of a real number.

${\left(x - 5\right)}^{2} = 0$ when $x - 5 = 0$, that is when $x = 5$.

Hence the vertex is at $\left(5 , 0\right)$.

The axis is vertical, passing through the vertex, so its equation is $x = 5$.

The intercept with the $y$ axis occurs when $x = 0$, so substitute $x = 0$ into the equation to find $y = {\left(0 - 5\right)}^{2} = {\left(- 5\right)}^{2} = 25$

So the intercept is at $\left(0 , 25\right)$

graph{(x-5)^2 [-39.58, 40.42, -5.28, 34.72]}