Question

Determine a quadratic function f(x)= ax^2 + bx + c if its graph passes through the point (2,19) and it has a horizontal tangent at (-1, -8)

Answer 1

After substituting both points into the formula you get the following:

19 = -4a + 2b + c
-8 = -a -b + c

Solving for c, we get:
c = a + b - 8

If we substitute c back into the first equation above, we get:
19 = -4a + 2b + (a + b - 8)
19 = -3a +3b - 8
27 = -3a + 3b
27+3a = 3b
(27+3a)/3 = b
9 + a = b

When we take the derivative and set f'(x) = 0, we end up with:
x = -b/2a
Now we substitute -1 for x, since it is the x vertex of the quad function along with b = (9 + a):
-1 = -(9+a)/2a
-1 = (-9-a)/2a
-2a = -9-a
9+a = 2a
9 = a

So:
9 = a
18 = b

Now we do actually have enough information to find c:
Going back to the first formula:
-19 = -4(9) + 2(18) + c
19 = -36 + 36 + c
19 = c

Therefore, the original function is f(x) = -9x^2 + 18x + 19