FORM QUAD. FUNCTIONS WITH THE FOLLOWING PROPERTIES...?

We don't know that the vertex is f(3) = -2, only that it's a point.

So we can start with the generic form of a quadratic:

y = ax² + bx + c

We are given three points:

(2, 0), (4, 0), and (3, -2)

If we substitute these into the above and simplify we get a system of three equations and three unknowns:

0 = a(2)² + b(2) + c and 0 = a(4)² + b(4) + c and -2 = a(3)² + b(3) + c

0 = 4a + 2b + c and 0 = 16a + 4b + c and -2 = 9a + 3b + c

I'll solve the first equation for c in terms of a and b then substitute into the other two:

0 = 4a + 2b + c

-4a - 2b = c

0 = 16a + 4b + c and -2 = 9a + 3b + c

0 = 16a + 4b - 4a - 2b and -2 = 9a + 3b - 4a - 2b

0 = 12a + 2b and -2 = 5a + b

Simplify the first by dividing both sides by 2, then solving the first for b in terms of a and substitute into the other equation:

0 = 12a + 2b

0 = 6a + b

-6a = b

-2 = 5a + b

-2 = 5a - 6a

-2 = -a

2 = a

Now we can solve for b and c:

b = -6a

b = -6(2)

b = -12

c = -4a - 2b

c = -4(2) - 2(-12)

c = -8 + 24

c = 16

Your quadratic is:

y = ax² + bx + c

y = 2x² - 12x + 16

There are roots at x = 2 and x = 4.

The vertex is at (3, -2)

Vertex Form:

 y  = a(x - 3)^2 - 2

Plug in one of the roots to find a.

0 = a (2 - 3)^2 - 2

0 = a - 2

a = 2

y = 2(x - 3)^2 - 2