Finding the Intercepts and Asymptote of a Graph with Exponential Functions

[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.

 

[Bohrok]

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?

 

[TyErd]

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

 

[TyErd]

not sure how to find b

 

[vela]

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

 

[Bohrok]

Think of what happens with y = e^2x - 3ke^x as x→∞ or -∞. Does it approach a certain value? What happens when you shift the graph up 5 units with the +5?

 

[Mark44]

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

Think of what happens with y = e^2x - 3ke^x as x→∞ or -∞. Does it approach a certain value? What happens when you shift the graph up 5 units with the +5?

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

 

[vela]

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

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

 

[Mark44]

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

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

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.

 

[TyErd]

hey I am still having trouble understanding what the horizontal asymptote is.

 

[TyErd]

b is 5 but i don't get why

 

[vela]

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

 

[TyErd]

do i have to use the calculator to do that?

 

[Altabeh]

do i have to use the calculator to do that?

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.

 

[Mark44]

Of course not!
Use the fact that for any x, x^a =0 if a-->-oo.

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.

 

[TyErd]

ook,as x approaches -infinity it gets closer to 5 but at a certain -x value, y actually equals 5. Which isn't right because its suppose to be an asymptote. And also why exclude positive infinity??

 

[Mark44]

ook,as x approaches -infinity it gets closer to 5 but at a certain -x value, y actually equals 5. Which isn't right because its suppose to be an asymptote. And also why exclude positive infinity??

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 = e^2x - 6e^x + 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.

 

[TyErd]

Im not sure if I am 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.

 

[vela]

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 $3\times10^{-434}$.

 

[TyErd]

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