SOLUTION: The equation f(x) = ac^x+d can be used to represent the following function. Determine the values of a, c, and d algebraically. Show all your work. Please view graph here;

Algebra ->  Functions -> SOLUTION: The equation f(x) = ac^x+d can be used to represent the following function. Determine the values of a, c, and d algebraically. Show all your work. Please view graph here;       Log On


   



Question 1180810: The equation f(x) = ac^x+d can be used to represent the following function. Determine the values of a, c, and d algebraically. Show all your work.
Please view graph here;
https://docs.google.com/document/d/1Cl3HrcoR6jibNzxfDTIilS3imSplLY8nQUU0OpU9vLA/edit?usp=sharing

Found 2 solutions by greenestamps, MathLover1:
Answer by greenestamps(13195) About Me  (Show Source):
You can put this solution on YOUR website!


f(x)=ac^x+d

We need to determine the values of three unknowns, so we need three equations. We can get them using three carefully chosen points on the graph.

(1) (-infinity, 4)

As x goes to negative infinity, ac^x goes to 0.
ac^x+d = 0+d = d = 4

(2) (0,1)

ac^x+d = ac^0+d = a(1)+d = a+4 = 1
a = -3

(3) (1,-2)

ac^x+d = ac+d = -2
-3c+4 = -2
-3c = -6
c = 2

ANSWER: f%28x%29=-3%282%5Ex%29%2B4

A graph....

graph%28400%2C400%2C-6%2C3%2C-15%2C6%2C-3%282%5Ex%29%2B4%29


Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

use these points on a graph:
(0,1)
(-5,4)
(1,-2)
set the system to determine the values of a, c, and d+
(0,1)

f%28x%29%E2%80%89=+ac%5Ex%2Bd
1=+ac%5E0%2Bd+........solve for a
1=+a%2A1%2Bd+
1=+a%2Bd+
a=1-d ....eq.1
(-5,4)
4=+ac%5E-5%2Bd+
4=+a%281%2Fc%5E5%29%2Bd+......solve for a
a=%284-d+%29%2F%281%2Fc%5E5%29
a=%284-d+%29c%5E5....eq.2

(1,-2)
-2=+ac%5E1%2Bd+......solve for a
-2=+ac%2Bd+
-2-d=+ac
a=%28-2-d%29%2Fc ....eq.3

from eq.1 and eq.2 we have
1-d+=%284-d+%29c%5E5.........solve for c
%281-d%29+%2F%284-d+%29=c%5E5
c=%28%281-d%29+%2F%284-d+%29%29%5E%281%2F5%29...................1)

from eq.1 and eq.3 we have
1-d+=%28-2-d%29%2Fc
c%281-d+%29=%28-2-d%29............solve for c
c=%28-2-d%29%2F%281-d+%29.........................2)

from 1) and 2) we have
%28%281-d%29+%2F%284-d+%29%29%5E%281%2F5%29=%28-2-d%29%2F%281-d+%29.........solve for d
d=4.10576.......round
d=4
go to
a=1-d ....eq.1, substitute d
a=1-4
a=-3
go to
c=%28-2-d%29%2F%281-d+%29.........................2), substitute d+
c=%28-2-4%29%2F%281-4+%29
c=2


f%28x%29%E2%80%89=+-3%282%29%5Ex%2B4

+graph%28+600%2C+600%2C+-7%2C+7%2C+-7%2C+7%2C-3%282%29%5Ex%2B4%29+