1. ## Need help with writing a quadratic equation

I am given the following info:

2 x-intercepts: (3,0) (9,0)
there is a maximum occurring at 2

I figured I would use the equation y=a(x-p)(x-q)

so I plugged in the two x-ints

y=a(x-3)(x-9)

Now my question is that if I know that the maximum is 2, where exactly would that go in the equation?

2. Originally Posted by coolieman
I am given the following info:

2 x-intercepts: (3,0) (9,0)
there is a maximum occurring at 2

I figured I would use the equation y=a(x-p)(x-q)

so I plugged in the two x-ints

y=a(x-3)(x-9)

Now my question is that if I know that the maximum is 2, where exactly would that go in the equation?
Hello,

you've got the 2 zeros of the quadratic function. That means the vertex of the corresponding parabola is at $x_V\left(\frac{3+9}{2}, a(6-3)(6-9) \right)$.

If I understand your problem correctly the maximum is 2 and that is only possible if this is the y-value of the vertex. Thus:

$a(6-3)(6-9) = 2~\Longrightarrow~-9a = 2~\Longrightarrow~a = -\frac{2}{9}$

The complete equation of the quadratic function is now:

$p(x)=-\frac{2}{9}(x-3)(x-9) =\boxed{ -\frac{2}{9}x^2+\frac{8}{3}x - 6}$