Y = 3X^2 + K
Vertex Form:
Y = a(X -h)^2 + k
= 3(X -0)^2 + K
Y = 3(X-2)(X+2),
= 3(X^2 -4)
= 3X^2 - 12
h = -b / 2a = 0 / 6 = 0.
K = Yv = 3*0^2 -12 = -12. V(0 , -12).
Let's use a positive 2 in the 1st
parenthesis of our factored Eq.
Y = 3(X + 2 )(X + 2) = 3(X + 2)^2 =
3(X^2 + 4X + 4) = 3X^2 + 12X + 12.
h = Xv = -b / 2a = -12 / 6 = -2.
This value of h does not agree with
our given Eq in which h = 0.
Therefore, -2 is the correct choice.
For what real number k does 3x^2 � k factor as
3(x � 2)(x + 2)?
Can someone please help with this question? I can not understand how it says to solve this in the book
1 answer