# How do you graph y=x^2- 6x+2?

May 22, 2018

$\text{y-int} = 2$
$\text{x-int} = 3 \pm \sqrt{7}$
vertex = $\left(3 , - 7\right)$

#### Explanation:

$y = {x}^{2} - 6 x + 2$

your function is in the form: $a {x}^{2} + b x + c$

$a = 1$
$b = - 6$
$c = 2$

From this we can derive:

y-intercept = c

y-int = 2

x-intercept(s) if they exist are the solutions or roots:

$y = {x}^{2} - 6 x + 2$

Roots (found with the quadratic formula): $3 \pm \sqrt{7}$

axis of symmetry is: $a o s = \frac{- b}{2 a} = \frac{- \left(- 6\right)}{2 \cdot 1} = 3$

vertex is: $\left(a o s , \text{*} f \left(a o s\right)\right)$
*this means you put the aos value back into the function as x and solve for y.

vertex is: $\left(3 , {3}^{2} - 6 \cdot 3 + 2\right)$

vertex is: $\left(3 , - 7\right)$