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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.
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?
oooohk so i found k and it equals 2 and also a which equals loge(5). and no i haven't found b
If z=e^x, you have y=z^2-6z+5. Does that help?
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?
If z=e^x, you have y=z^2-6z+5. Does that help?
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?
vela is leading you toward finding the x-intercepts, and Bohrok is leading you toward finding the horizontal asymptote.
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.
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.
hey im still having trouble understanding what the horizontal asymptote is.
b is 5 but i dont get why
What does the function do as x goes to infinity or -infinity?
do i have to use the calculator to do that?
Altabeh
Jan14-10, 07:57 AM
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.
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.
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??
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??
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.
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.
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}.
oh yea vela you're right, i just tried it on a different calculator.
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