# How do you find the vertex and intercepts for y = (x-5)^2 +2?

Apr 18, 2016

(5,2) , (0,27) , no x-intercepts

#### Explanation:

The standard vertex form of a quadratic function is

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{y = a {\left(x - h\right)}^{2} + k} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
where (h , k) are the coordinates of the vertex , and a is a constant.

the function $y = {\left(x - 5\right)}^{2} + 2 \text{ is in this form }$

by comparison, the coords of vertex = (5 , 2)

To find where it crosses the y-axis , let x = 0 in the equation.

x = 0 : y = ${\left(- 5\right)}^{2} + 2 = 25 + 2 = 27 \Rightarrow \left(0 , 27\right)$

To find where it crosses the x-axis let y = 0

y $= 0 : {\left(x - 5\right)}^{2} + 2 = 0$

hence (x-5)^2 = -2 rArr x-5 = ±sqrt-2

rArr x = 5 ± 2i " no real solution thus no x-intercepts "
graph{(x-5)^2+2 [-40, 40, -20, 20]}