How do you find the x intercepts and vertex of #y=x^2-2x-4#?
1 Answer
Dec 15, 2016
X-intercepts are
#(1-sqrt5, 0)#
#(1+sqrt5, 0)#
Vertex
Explanation:
Given -
#y=x^2-2x-4#
To find the x-intercepts put
#x^2-2x-4=0#
#x^2-2x=4#
#x^2-2x+1=4+1#
#(x-1)^2=5#
#x-1=+-sqrt5#
X-intercepts are
#(1-sqrt5, 0)#
#(1+sqrt5, 0)#
Vertex
#x=(-b)/(2a)=(-(-2))/(2xx1)=2/2=1#
At
#y=(1)^2-2(1)-4=1-2-4=-5#
Vertex
