SOLUTION: A quadratic function has x-intercepts at (-8,0) and (2,0). What is the transformational form of the function if the minimum y value is -3? a) y=3/25(x+3)^2-3 b) 25/3(y+3)=(x+3)^2

Question 121328: A quadratic function has x-intercepts at (-8,0) and (2,0). What is the transformational form of the function if the minimum y value is -3?
a) y=3/25(x+3)^2-3
b) 25/3(y+3)=(x+3)^2
c) 25/3(y-3)=(x-3)^2
d) 3/25(y-3)=(x-3)^2
Answer by Edwin McCravy(20056) (Show Source): You can put this solution on YOUR website!
A quadratic function has x-intercepts at (-8,0) and (2,0).
What is the transformational form of the function if the
minimum y value is -3?
a) y=3/25(x+3)^2-3
b) 25/3(y+3)=(x+3)^2
c) 25/3(y-3)=(x-3)^2
d) 3/25(y-3)=(x-3)^2


Since it has intercepts (-8,0) and (2,0), the axis of
symmetry must pass through their midpoint, which is 

(,) or (,) = (-3,0)

The axis of symmetry has equation x = h, which is x = -3 

The vertex then must be (h,k) where h = -3 and k is the 
minimum y value, which is -3.

The equation of a parabola is

   c(y-k) = (x-h)^2

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

   c(y+3) = (x+3)^2

Since the parabola passes through (2,0)

   c(0+3) = (2+3)^2
       3c = 5^2
       3c = 25
        c = 25/3

So  

   c(y+3) = (x+3)^2

is now

   (y+3) = (x+3)^2

So the answer is (b)

Edwin

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