# How do you write y=x^2-6x+8 in vertex form and identify the vertex, y intercept and x intercept?

May 23, 2015

$y = {x}^{2} - 6 x + 8$

x of vertex: $x = \left(- \frac{b}{2} a\right) = \frac{6}{2} = 3$
y of vertex: y = f(3) = 9 - 18 + 8 = -1

Vertex form:$y = {\left(x - 3\right)}^{2} - 1$

y-intercept when x = 0 --> y = 8

x-intercept when y = 0, solve ${x}^{2} - 6 x + 8 = 0.$

Compose factor pairs of 8 -> (1, 8)(2, 4). This sum is 6 = -b. Then the 2 real roots (x-intercepts) are 2 and 4