2006-10-26 17:33:21 · 5 answers · asked by Steven B 1 in Science & Mathematics ➔ Mathematics
It's an exponential curve but it starts below the x-axis. It also crosses at (0,-1 7/8) and (4,0). It asymptotically approaches y = -2 at negative infinity.
I've attached a picture that should give you an idea of the actual graph...
2006-10-26 17:50:34 · answer #1 · answered by Puzzling 7 · 3⤊ 0⤋
You have two options. You can buy a graphing calculator and put this equation in to graph or you can plug in numbers for x and graph them on graphing paper. I will show you some sample points to graph:
f(5)=2^(5-3)-2
f(5)=2^2-2
f(5)=4-2
f(5)=2
(5,2)
f(6)=2^(6-3)-2
f(6)=2^3-2
f(6)=8-2
f(6)=6
(6,6)
f(0)=2^(0-3)-2
f(0)=2^-3-2
f(0)=1/8-2
f(0)=1/8-16/8
f(0)=-15/8
(0,-15/8)
With those points, you should be able to make a graph out of it.
2006-10-27 00:58:01 · answer #2 · answered by Anonymous · 2⤊ 0⤋
first you need to add 2 to both sides. Then you need to add a log or an ln to both sides. with the logs added, then you bring the exponent down in front on the log. Then solve for x and solve for f(x) using multiple points. it should cross the y axis around -2 and swoop upward and cross the x axis around 4.
2006-10-27 00:55:20 · answer #3 · answered by Dawn J 4 · 3⤊ 0⤋
check ur e-mail
2006-10-27 01:15:31 · answer #4 · answered by M. Abuhelwa 5 · 1⤊ 0⤋
Sorry I don't have time to do your homework tonight.
2006-10-27 00:41:40 · answer #5 · answered by The Count 4 · 0⤊ 5⤋