find the height, in feet of the rock after 10 seconds

Using the form: ax^2 + bx
+ c = y
y = h(x), c = 0
equation 1
x=1; y=58
a(1^2) + 1b = 58
a + b = 58
equation 2
x=2; y=112
a(2^2) + 2b = 112
4a + 2b = 112
:
Multiply the 1st equation by 2, subtract from the 2nd equation
a + b = 58 x 2 = 2a+2b = 116
subtract that from 2nd
equation
4a +2b =112 – 2a+2b= 116 =
2a = -4
A = -4/2 = -2
Find b using: a + b = 58
-2 + b = 58
b = 58 + 2
b = 60
use equation: h(x) = -2x^2 + 60x
Find the height, in feet,of the rock after 10 seconds in the air.;
x=10
h(10) = -2(10^2) + 60(10)
h(10) = -2(100) + 600
h(10) = -200 + 600
h(10) = 400 ft
The rock is 400 feet in the air after 10 seconds