algebra Please Help
posted by Jennifer .
New question: A baseball was 2.74 feet above ground when it was hit. It reached a max. height of 116.3ft when it was approx 215.3 ft away from where he hit the ball. The ball lands after travelling a ground distance of approx 433.1 ft.
Find an equation of form y = A(xh)^2 + k where vertex is (h,k) and constant A is a scaling factor.
So the vertex is (215.3, 116.3).
Plugging in I have
y = A (x215.3)^2 + 116.3
Using point (0, 2.74), I have
2.74 = A (0215.3)^2 + 116.3
2.74 = A (46354.09 + 116.3)
2.74 = A 46,470.39
0.000059 = A
**** shouldn't my A be negative to get a hill shape?

Yes. You made a mistake, placing the right parenth ) the right of the 116.3.
A will be negative if you solve
46354 A = 2.74  116.3 = 113.56
A = 0.0024498
Check: when x = 433.1,
y = 0.0024498(433.1215.3)^2+116.3
= 116.2 +116.3 = 0.01
That's close enough.