# How do you identify the important parts of y = (x – 1)^2 – 16 to graph it?

Sep 26, 2015

Minimum point $\Rightarrow$ $\left(1 , - 16\right)$
$y -$intercept $\Rightarrow$ $\left(0 , - 15\right)$

#### Explanation:

From the completed square, you already know that the minimum points are $\left(1 , - 16\right)$

To find the $y -$intercept, you need to expand the brackets to get it in the form of $a {x}^{2} + b x + c$ , where $c$ is your $y -$intercept.

So,

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

So $15$ is your $y -$intercept.

graph{(x-1)^2-16 [-9.495, 10.505, -17.76, -7.76]}