# hard time with graph problems

• Jun 11th 2007, 05:53 PM
wvmcanelly@cableone.net
hard time with graph problems
Identify the exponential function of the form
f (x) = a(2 x ) + b whose
graph is shown in the figure.
a. f (x) = 3(2 x )
b.

f (x) = 2 x 3
c.

f (x) = 2(2 x ) 4
d. f (x) = 2 x 2

The graph is attached a a work file.

thanks for any help

• Jun 11th 2007, 06:18 PM
Jhevon
Quote:

Originally Posted by wvmcanelly@cableone.net
Identify the exponential function of the form
f (x) = a(2 x ) + b whose
graph is shown in the figure.
a. f (x) = 3(2 x )
b.

f (x) = 2 x 3
c.

f (x) = 2(2 x ) 4
d. f (x) = 2 x 2

The graph is attached a a work file.

thanks for any help

what you have in brackets, are those suppsoed to be powers? when you say 3(2 x) do you mean $\displaystyle 3 \cdot 2^x \mbox { or } 3^{2x}$ ?
• Jun 11th 2007, 06:24 PM
wvmcanelly@cableone.net
My apologies,

I'm half asleep.

f(x) = 3(2^x)
f(x) = 2^x-3
f(x) = 2(2^x) - 4
f(x) = 2^x - 2
• Jun 11th 2007, 06:32 PM
Jhevon
Quote:

Originally Posted by wvmcanelly@cableone.net
My apologies,

I'm half asleep.

f(x) = 3(2^x)
f(x) = 2^x-3
f(x) = 2(2^x) - 4
f(x) = 2^x - 2

now we could go into all the theories about how to find asymptotes and how an exponential function behaves and whatnot, but here, i think it is simplest to just check the points given

on the graph, we have the points (0, -2) and (2, 1). these points will work for only one of the choices.

(0, - 2) means, when x is 0, y is -2, so let's test that

plug in x = 0 in the first, we get y = 3, so thats not our guy, so (a) is out!

plug in x = 0 in the second, we get y = -2, so this MAY be our guy, (b) is still in

plug in x = 0 in the third, we get y = -2, so (c) is still in

plug in x = 0 in the fourth, we get y = -1, so (d) is out!

Now just check (b) and (c) for the second point.

(2,1) means, when x = 2, y = 1

so for (b), plug in x = 2, we get y = 1, ah, this seems to be it!

so for (c), plug in x = 2, we get y = 4, this is not our guy!

so the answer is (c), y = 2^x - 3

NOTE: we could also realize that 2^x has a horizontal asymptote at y = 0. adding -3 shifts eveything down 3 units, so the asymptote becomes y = -3. no other graph is like that, so we could have gotten the answer that way
• Jun 12th 2007, 08:44 AM
wvmcanelly@cableone.net
Thanks, I really have a hard time with graphs. I was really stuck on this one and I appreciate you explaining the problem