# How do you find the x intercepts for the parabola and a vertex y = x^2 – 6x + 5?

Apr 15, 2015

Your quadratic is in the form $y = a {x}^{2} + b x + c$ where:
$a = 1$
$b = - 6$
$c = 5$

To find the $x$ intercept(s) set $y = 0$ and solve for $x$ the 2nd degree equation: ${x}^{2} - 6 x + 5 = 0$

The coordinates of the vertex are:
${x}_{v} = - \frac{b}{2 a}$
${y}_{v} = - \frac{\Delta}{4 a}$
Where $\Delta = {b}^{2} - 4 a c$