
Quadratic Question
Find the quadratic function f such that f (2) = f (4) = 0 and 7 is the greatest value of f (x).
I know that
$\displaystyle k(x2)(x4) $ because the parabola is a negative one because it has a greatest value and because the x ints are 2 and 4
I don't know what to do from here


A better solution:
Your $\displaystyle \displaystyle x$ intercepts are at $\displaystyle \displaystyle x = 2$ and $\displaystyle \displaystyle x = 4$, so the axis of symmetry is $\displaystyle \displaystyle x = \frac{2+4}{2} = 3$ So the turning point is $\displaystyle \displaystyle (3, 7)$.
Therefore the equation of the parabola is:
$\displaystyle \displaystyle y = a(x  3)^2 + 7$.
Now substituting another of your points that lie on the curve, e.g. $\displaystyle \displaystyle (2, 0)$, you find
$\displaystyle \displaystyle 0 = a(2  3)^2 + 7$
$\displaystyle \displaystyle 0 = a + 7$
$\displaystyle \displaystyle a = 7$.
So the equation of your parabola is:
$\displaystyle \displaystyle y = 7(x  3)^2 + 7$.

Yet another way to do this if you don't happen to know offhand that the vertex is always exactly midway between the xintercepts: $\displaystyle k(x 2)(x 4)= k(x^2 6x+ 8)$. Now complete the square to find the highest point:
$\displaystyle k(x^2 6x+ 8)= k(x^2 6x+ 9 9+ 8= k((x 3)^2 1)$
Since a square is never negative $\displaystyle (x 3)^2 1$ will have its lowest point (and $\displaystyle k((x3)^2 1)$ its highest) when, just as Prove It said, when x= 3. At that point $\displaystyle k((3 3)^2 1)= k= 7$.