SOLUTION: identify the exponential function of the form f(x)=a(2^x)+b whose graph is shown in figure. info: 2 points on graph (0,-2), (2,1) y=-3

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: identify the exponential function of the form f(x)=a(2^x)+b whose graph is shown in figure. info: 2 points on graph (0,-2), (2,1) y=-3       Log On


   



Question 63196: identify the exponential function of the form f(x)=a(2^x)+b whose graph is shown in figure.
info:
2 points on graph (0,-2), (2,1)
y=-3


Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
identify the exponential function of the form
f(x) = a(2^x) + b
:
write it:
a(2^x) + b = y
:
2 points on graph (0,-2), (2,1)
Substitute 0 for x and -2 for y:
a(2^0) + b = -2
a(1) + b = -2
a + b = -2
:
Do the same for x=2, y=1
a(2^2) + b = 1
4a + b = 1
:
Use elimination on these two equations to find a:
4a + b = 1
a + b = -2
------------- subtract eliminating b
3a = 3
a = 1
:
Find b using 4a + b =1
4(1) + b = 1
4 + b = 1
b = 1 - 4
b = -3
:
Our equation is f(x) = (2^x) - 3
Graph of this:
+graph%28+300%2C+200%2C+-10%2C+10%2C+-10%2C+10%2C+%282%5Ex%29+-+3+%29+
y=-3: If you look closely you can see that y approaches -3 but does not reach it, as you calculate neg x values of 2^x