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

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Answer 1 · 2015-04-15T19:44:55

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

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

The coordinates of the vertex are:
$x_v=-b/(2a)$
$y_v=-Delta/(4a)$
Where $Delta=b^2-4ac$

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