How do you graph #y= [-2 (x - 3)] ^2#?

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Answer 1 · 2018-05-22T14:35:34

see graph below.

First square it out:

#y =[-2 (x - 3)] ^2#

#y=(-2)^2 (x - 3)^2#

#y=4(x-3)^2#

#y=4(x^2-6x +9)#

#y=4x^2-24x +36#

now you have it in standard form #ax^2 + bx + c#

a=4
b=-24
c=36

c is your y-intercept #y=36#

axis of symmetry (aos) is: #(-b)/(2a) = (-(-24))/(2*4) = 24/8 =3#

vertex #(h,k) = (aos, f(aos)) = (3, (4*3)^2-24*3 +36) = (3,0)#

x-intercepts are the roots:

#y=4(x-3)^2#

#x = 3#

graph{4x^2-24x +36 [-9.71, 10.29, -1.6, 8.4]}