Very simple.
Let's say you have an equation.
f(x) = x^2
You are asked to find the value for y when x equals 1.
The new equation is: f(1) = (1)^2
f(1) = 1
When x = 1, y = 1.
The same concept is applied here.
In the graph, where does x equal 0?
It equals zero at the origin.
Is there any y-value associated with 0?
Yes, there is.
Y equals five when x equals 0.
So
h(0) = 5
A event would be five I think and the event of b would be nine
(0,-1)(1,0)(2,3)(3,8)(4,15)
Exponential growth has the form:
F=Ir^t, F=final amount, I=initial amount, r=rate, t=time.
First let's solve for the rate, r, using the points provided...
37325/34560=(Ir^2)/(Ir^1)
r≈1.08
Now find the initial amount...
34560=I(1.08)^1
I=32000 so now we have the entire equation, using x and y...
y=32000(1.08)^x
The answer is 272 + 102 which equals 374yd^2
