How do you find the vertex and the intercepts for #y= x^2 - 4x + 5#?
1 Answer
Apr 6, 2017
Vertex
Y-intercept
Explanation:
Given -
#y=x^2-4x+5#
#x=(-b)/(2a)=(-(-4))/(2xx 1)=4/2=2#
At
#y=2^2-4(2)+5=4-8+5=9-8=1#
Vertex
At
#y=0^2-4(0)+5=0-0+5=5#
#y=5#
Y-intercept#(0, 5)#
