How do you write #y = 4(x-3)^2+1 # in standard form?
1 Answer
Jul 2, 2018
#y=4x^2-24x+37#
Explanation:
Given -
#y=4(x-3)^2+1#
#y=4(x^2-6x+9)+1#
#y=4x^2-24x+36+1#
#y=4x^2-24x+37#
