• Jan 15th 2011, 04:04 AM
jgv115
Find the quadratic function f such that f (2) = f (4) = 0 and 7 is the greatest value of f (x).

I know that

$\displaystyle -k(x-2)(x-4)$ 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
• Jan 15th 2011, 04:08 AM
Also sprach Zarathustra
k=7 and you done!
• Jan 15th 2011, 04:32 AM
Prove It
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:

$\displaystyle \displaystyle y = -7(x - 3)^2 + 7$.
• Jan 15th 2011, 07:02 AM
HallsofIvy
Yet another way to do this if you don't happen to know off-hand that the vertex is always exactly midway between the x-intercepts: $\displaystyle -k(x- 2)(x- 4)= -k(x^2- 6x+ 8)$. 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 $\displaystyle -k((x-3)^2- 1)$ its highest) when, just as Prove It said, when x= 3. At that point $\displaystyle -k((3- 3)^2- 1)= k= 7$.