How do you graph #f(x)=(1/2)^x -3# by plotting points?

1 Answer
Apr 27, 2017

Pick values for #x# that are easy to solve

Explanation:

So we can choose the following #x# values:
#x=-3,-2,-1,0,1,2#

Plugging in the values we get the following
#f(-3)=\frac{2^3}{1^3}-3=2^3-3=8-3=5#
#f(-2)=\frac{2^2}{1^2}-3=2^2-3=4-3=1#
#f(-1)=\frac{2^1}{1^1}-3=2^1-3=2-3=-1#
#f(0)=\frac{1^0}{2^0}-3=1-3=-2#
#f(1)=\frac{1^1}{2^1}-3=\frac{1}{2}-3=-\frac{5}{2}#
#f(2)=\frac{1^2}{2^2}-3=\frac{1}{4}-3=-\frac{11}{4}#

You can plug in the points #(-3,5),(-2,1),(-1,-1),(0,-2),(1,-2.5),(2,-2.75)# to get
graph{(1/2)^x-3 [-14.85, 10.46, -3.99, 8.67]}