# How do you find the vertex and the intercepts for y=x^2-10x+20?

Vertex is at $\left(5 , - 5\right)$ y-intercept is at $\left(0 , 20\right)$ x-intercepts are at $\left(2.764 , 0\right) \mathmr{and} \left(7.236 , 0\right)$
$y = {x}^{2} - 10 x + 20 = {\left(x - 5\right)}^{2} - 25 + 20 = {\left(x - 5\right)}^{2} - 5 \therefore$Comparing with vertex form of equation; $y = a {\left(x - h\right)}^{2} + k$ we get vertex at $\left(h , k\right) = \left(5 , - 5\right)$.To get y-intercept putting x=0 in the equation we get $y = 20$ and to find x-intercepts putting y=0 in the equation we get ${\left(x - 5\right)}^{2} - 5 = 0 \mathmr{and} {\left(x - 5\right)}^{2} = 5 \mathmr{and} \left(x - 5\right) = \pm \sqrt{5} \mathmr{and} x = 5 \pm \sqrt{5} \mathmr{and} x = 2.764 , 7.236 \therefore$Vertex is at $\left(5 , - 5\right)$ y-intercept is at $\left(0 , 20\right)$ x-intercepts are at $\left(2.764 , 0\right) \mathmr{and} \left(7.236 , 0\right)$ graph{x^2-10x+20 [-40, 40, -20, 20]}[Ans]