Form quadratic equation with given info...?

You want a quadratic that has an f(x) greater than 0 between 1 < x < 5.  This means it would open downwards so the leading coefficient is negative.

In order for f(x) to be greater than 0 when 1 < x < 5, then that means that f(x) = 0 at x = 1 and 5.  So we can start there.

If x = 1 and 5 are the roots, then (x - 1) and (x - 5) are the factors.  We can multiply these numbers together, set them equal to 0, then multiply both sides by an unknown "a".  We get:

0 = (x - 1)(x - 5)

a0 = a(x - 1)(x - 5)

0 = a(x - 1)(x - 5)

Now let's expand the left side:

0 = a(x² - 6x + 5)

0 = ax² - 6ax + 5a

Since we want there to be a maximum we know a must be negative.

We want that maximum to be 8.  If we turn our above expression into a function we can then put this into vertex form: 

y = a(x - h)² + k

Where "k" is the maximum value.  We can set that last term from the expression to 8 and solve for the remaining unknown.

Now we have:

y = ax² - 6ax + 5a

We want to complete the square so we need the right side to be in the form of (x² + bx).  We can subtract 5a from both sides then divide both sides by a:

y - 5a = ax² - 6ax

(y - 5a) / a = x² - 6x

Now we can complete the square by adding 9 to both sides:

(y - 5a) / a + 9 = x² - 6x + 9

Factor the right side then solve for y again, starting with subtracting 9 from both sides, multiplying both sides by a, then adding 5a to both sides:

(y - 5a) / a + 9 = (x - 3)²

(y - 5a) / a = (x - 3)² - 9

y - 5a = a(x - 3)² - 9a

y = a(x - 3)² - 4a

We want the maximum to be 8 so we can set that last term equal to 8 and solve for "a":

-4a = 8

a = -2

Now that we know "a" we can find out original quadratic:

y = ax² - 6ax + 5a

y = -2x² + 12x - 10