How do you find the vertex and intercepts for #y = (x-5)^2 +2#?
1 Answer
Apr 18, 2016
(5,2) , (0,27) , no x-intercepts
Explanation:
The standard vertex form of a quadratic function is
#color(red)(|bar(ul(color(white)(a/a)color(black)( y =a (x - h)^2 + k)color(white)(a/a)|)))#
where (h , k) are the coordinates of the vertex , and a is a constant.the function
#y = (x - 5)^2 + 2 " 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 =
#(-5)^2 + 2 = 25 + 2 = 27 rArr (0 , 27) # To find where it crosses the x-axis let y = 0
y
#= 0 : (x-5)^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]}
