# How do you write y=x^2-2x+8 in vertex form and identify the vertex, y intercept and x intercept?

May 19, 2015

$y = {x}^{2} - 2 x + 8$

x of vertex: $x = \left(- \frac{b}{2} a\right) = \frac{2}{2}$= 1#

y of vertex: $f \left(1\right) = 1 - 2 + 8 = 7$

Vertex form: $y = {\left(x - 1\right)}^{2} + 7.$

Find x-intercepts by solving quadratic equation:
$y = {x}^{2} - 2 x + 8 = 0$.
Use the new AC Method to solve quadratic equation. Compose factor pairs of c = 8: (-2, 4). This sum is -2 + 4 = 2 = -b. Then the 2 real roots (x -intercepts) are: -2 and 4.