The figure shows the graph of f(x)= e ^x Use transformations of this graph to graph each function. "Be sure to give equations of the asymptotes"
( points are -2, .14 -1, .37 0,1 1, 2.72 2, 7.39)



Given equation to graph and find asymptotes :
h(x)= e^(x+1) - 1


I graphed it... just down 1 from each y and left 1 from each x... but how do I find the asymptotes?

Thanks

the only asymptote in the basic f(x)=e^x function is a horizontal asymptote along the line y=0
So the only shifts that should affect your asymptote are vertical shifts.
Knowing that, determine your vertical shift and go from there!

how do you know it does not have a vertical asymptote? how did you get y=0?

so given the new equation the asymptote should be horizontal and it is -1? would i say -1 or y=-1?


they showed us tricks to figure them out where there are 3 cases n = d n >d and d > n but I dont see how I do it for this one? Do I just write it as e^x / e^(x+1) - 1 to get a numerator and denominator?

cuz the ones they showed us were things like x^2 + 1 / (over) x ^3 + 1

so its obvious to look at numerator and denominator and figure out what to do.

This post was edited by fingerling on Feb 16 2014 05:52pm

If it had a vertical asymptote, then the would be an x-value that you could plug in and not get an answer.

You would say there's a horizontal asymptote at y=-1

Those tricks you mentioned are for fractions. This function is not a fraction, so those can't be used.
Remember them, because they are very handy when dealing with fractions!

These are from the same direction set:

g(x) = e ^ x + 2

no vertical
horizontal = 2

G(x) = e^(x+1)
no vertical
horizontal = -1

yes?

re-check your 2nd one.
1st one looks ok

x +1 means we are shifting one spot on x axis to the left??? so -1?


also for this one

h(x) - e ^ (x+1) - 1

if x = -1 = e ^ 0 = 1 (-1) = 0

vertical asymptote = -1
horizontal = ????

i hate math :(((

what is the difference between the graphs of e^x and e^(x+1)?
yes, the whole graph is shifted - you got that part right
but what is the asymptote of e^x?

if you look at/imagine the graph, you see that it never crosses the x-axis but is always above
starting very close to the x-axis from -infinity, growing very slowly first and then accelerating
so the asymptote of e^x is the x-axis
does the shift produced by moving slightly to the left have any impact on the asymptote?
no - the x-axis is still the asymptote

no good hating math
embrace it, love it and it will get easier to deal with :))))))

this makes sense. thank you