SOLUTION: Basic question. f(x) equals y because plugging in some value x into the function will give you what the corresponding y-value will be. And the y-value of a LINEAR FUNCTION will alw

Algebra ->  Functions -> SOLUTION: Basic question. f(x) equals y because plugging in some value x into the function will give you what the corresponding y-value will be. And the y-value of a LINEAR FUNCTION will alw      Log On


   



Question 1001042: Basic question. f(x) equals y because plugging in some value x into the function will give you what the corresponding y-value will be. And the y-value of a LINEAR FUNCTION will always be the same thing as its horizontal asymptote?
I know the function will output differently for a rational function, and then the rules change. but for something like lim x -> ∞ where the function is f(x) = 5+e^(-x^2). By plugging in ∞ into the function this will output the horizontal asymptote because the horizontal asymptote is the same as y-value. In this case y = 5 and that is the horizontal asymptote.
not sure if this is right or not.
Please help!

Answer by ikleyn(52787) About Me  (Show Source):
You can put this solution on YOUR website!
.
Below is the plot of the functions  y1 = 5+%2B+e%5E%28-x%5E2%29  (in red)  and  y2 = e%5E%28-x%5E2%29  (in green)  with the horizontal asymptote  y = 5  (in blue).



Figure. Plots y1 = 5+%2B+e%5E%28-x%5E2%29 and y2 = e%5E%28-x%5E2%29

The function  y2 = e%5E%28-x%5E2%29  tends to zero at  x ---> +/-infinity.

Therefore,  the function  y1 = 5+%2B+e%5E%28-x%5E2%29  tends to  5 at  x ---> +/-infinity.

y = 5  is the horizontal asymptote for the function  y1.

Notice that the function  y1  is not a rational function,  as well as the function  y2.