How do you graph the parabola #Y=(x-4)^2# using vertex, intercepts and additional points?
1 Answer
Aug 3, 2017
Y-intercept #(0, 16)
x-intercept
Vertex
Explanation:
Given -
#y=(x-4)^2#
It's y- intercept at
#y=(0-4)^2=(-4)^2=16#
#(0, 16)#
Its x - intercept at
#(x-4)^2=0#
#x-4=0#
#x=4#
At
#(4,0)#
Vertex
Find the minimum point
#dy/dx=2(x-4)xx1#
#dy/dx=2x-8#
#dy/dx =0 => 2x-8=0#
#x=8/2=4#
X- coordinate of the vertex#x=4#
Its y-coordinate of the vertex is
Vertex
Vertex is the same as minimum point
For graph