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]}

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