Quadratic Equations

Question

A parabola has its maximum point at (-2, 11) and the y coordinate of the y-intercept is 5. 

What is the equation of the parabola?

(There is one aspect of this process which I am confused on. I am not looking for the answer, since I provide it, but only clarification where noted.)

f(x)=ax²+ bx +c

h=2 -b/2a= -2/1=-b b=4a

a(-2)²+ b(-2) +c

4a-2b+c=11

-4a-2(4a)+c=11

-4a-8a+c=11

-4a+c=11

-4a+5=11

-4a+6

a= -3/2

-3/2x²-6x+5 Where does (-6) come from? This is my question.

Answer 1

As you have the vertex it's easier to use the vertex form

a (x-h) ² + k

The vertex is (h,k) so h = -2, k = 11

So y = a (x+2)² + 11

x = 0, y = 5

So 5 = 4a + 11, giving a = -3/2

So equation is y = (-3/2) (x+2)² + 11

Answer 2

Hi Brian,

If you start with the equation y=a(x-h)²+k you should get your answer with the -6. Using h=-2 and k=11 and the y intercept of (0,5) you get y=-3/2(x+2)²+11. When you multiply out (x+2)² you get y=-3/2(x+4x+4)+11 and further simplification yields y=-3/2x²-6x-+5.

Let me know if you have anymore questions or if you need me to go in detail on one of the steps.

Thank you,

Mark P.

Answer 3

You figured out b = 4a. Then you figured out a was -3/2. 4 times -3/2 = -6.