To determine the exponential equation for the given graph, we can use the general form of an exponential function:
y = ab^x + c
Using the given points (0, 5), (1, 1), and (2, -11), we can substitute these values into the equation to determine the values of the base (b), coefficient (a), exponent, and constant (c).
1. Substituting (0, 5):
5 = ab^0 + c
5 = a + c
2. Substituting (1, 1):
1 = ab^1 + c
1 = ab + c
3. Substituting (2, -11):
-11 = ab^2 + c
We now have a system of three equations with three variables (a, b, c).
From equation 1, we have: a + c = 5 ----> Equation (4)
From equation 2, we have: ab + c = 1
Solving equations (4) and (2) simultaneously, we can eliminate c:
(a + c) - (ab + c) = 5 - 1
a - ab = 4 - 1
a(1 - b) = 3
a = 3 / (1 - b) ----> Equation (5)
Now, substitute the value of a from equation (5) into equation (1):
5 = (3 / (1 - b)) + c
Simplifying, we get:
5(1 - b) = 3 + c(1 - b)
5 - 5b = 3 + c - cb
Rearranging, we have:
2 = c - 5b + cb
2 = c(1 - 5b) ----> Equation (6)
Now, we have two equations (5) and (6) with two variables (b and c).
Equation (5) can be rewritten as:
a = 3 / (1 - b)
Substitute the value of a from equation (5) into equation (6):
2 = (3 / (1 - b))(1 - 5b)
Simplifying, we get:
2 = 3 - 15b - 3b^2
Rearranging, we have:
3b^2 + 15b - 1 = 0
Now, we can solve this quadratic equation for the value of b using the quadratic formula:
b = (-15 ± √(15^2 - 4 * 3 * (-1))) / (2 * 3)
Simplifying, we get:
b = (-15 ± √(225 + 12)) / 6
b = (-15 ± √(237)) / 6
The values of b obtained from this equation will be the base of the exponential equation.
Similarly, we can substitute the value of b into equation (5) to find the corresponding value of a, and substitute both b and a into equation (4) to find the value of c.
Once we have the values of a, b, and c, we can write the exponential equation in the form: y = ab^x + c.