# Homework Help: Graphic Exponentials help

1. Jan 13, 2010

### TyErd

The graph of a function with equation y=e^(2x) - 3ke^(x) +5 intersects the axes at (0.0) and (a,0) and has a horizontal asymptote at y=b. Find the exact values of a,b and k.

I tried assigning a variable to e^x but struck a dead end.

2. Jan 13, 2010

### Bohrok

Re: Exponentials

If (0, 0) is a point on the graph, then you can replace y and x with 0 to solve for k, and a won't be hard to solve for after that.
Have you found b?

3. Jan 13, 2010

### TyErd

Re: Exponentials

oooohk so i found k and it equals 2 and also a which equals loge(5). and no i haven't found b

4. Jan 13, 2010

### TyErd

Re: Exponentials

not sure how to find b

5. Jan 13, 2010

### vela

Staff Emeritus
Re: Exponentials

If $z=e^x$, you have $y=z^2-6z+5$. Does that help?

6. Jan 13, 2010

### Bohrok

Re: Exponentials

Think of what happens with y = e2x - 3kex as x→∞ or -∞. Does it approach a certain value? What happens when you shift the graph up 5 units with the +5?

7. Jan 13, 2010

### Staff: Mentor

Re: Exponentials

vela is leading you toward finding the x-intercepts, and Bohrok is leading you toward finding the horizontal asymptote.

8. Jan 14, 2010

### vela

Staff Emeritus
Re: Exponentials

Yeah, I misread the problem. Ignore what I said.

9. Jan 14, 2010

### Staff: Mentor

Re: Exponentials

No, not a problem. I was just pointing out to the OP that the two of you were looking at different parts of the problem.

10. Jan 14, 2010

### TyErd

Re: Exponentials

hey im still having trouble understanding what the horizontal asymptote is.

11. Jan 14, 2010

### TyErd

Re: Exponentials

b is 5 but i dont get why

12. Jan 14, 2010

### vela

Staff Emeritus
Re: Exponentials

What does the function do as x goes to infinity or -infinity?

13. Jan 14, 2010

### TyErd

Re: Exponentials

do i have to use the calculator to do that?

14. Jan 14, 2010

### Altabeh

Re: Exponentials

Of course not!

Just try to puzzle out where y goes for x-> -oo. I've excluded +oo to get you to understand something important. Use the fact that for any x, x^a =0 if a-->-oo.

15. Jan 14, 2010

### Staff: Mentor

Re: Exponentials

This not true for 0 < x < 1. Even if you limit it to x > 1, x^a merely approaches zero as a gets large. x^a is not equal to zero for any real value of a.

Last edited: Jan 14, 2010
16. Jan 14, 2010

### TyErd

Re: Exponentials

ook,as x approaches -infinity it gets closer to 5 but at a certain -x value, y actually equals 5. Which isnt right because its suppose to be an asymptote. And also why exclude positive infinity??

17. Jan 14, 2010

### Staff: Mentor

Re: Exponentials

as x approaches -infinity y approaches 5. The use of "it" is confusing because a casual reader would think you were talking about x. There is no negative value of x for which y = e2x - 6ex + 5 equals 5. In fact, for x < 0, y is always < 5, but it gets infinitesimally close to 5 the more negative x gets.

There is a positive value of x for which y = 5, but we're talking about asymptotic behavior for x < 0, i.e., behavior for very negative x values.

18. Jan 14, 2010

### TyErd

Re: Exponentials

Im not sure if im doing something wrong but I just substituted -15 into x and y equaled 5 but as i get more negative like x=-1000, y still equals 5.

19. Jan 14, 2010

### vela

Staff Emeritus
Re: Exponentials

You're probably running into the limitations of your calculator. For x=-15, to ten decimal places, y=4.9999981645, which is within two-millionths of 5. At x=-1000, y differs from 5 by about $3\times10^{-434}$.

20. Jan 14, 2010

### TyErd

Re: Exponentials

oh yea vela you're right, i just tried it on a different calculator.