Find the equation of the eventual asymptotes of the following functions.
If one overlays the lines y = 1 and x = -0.5 onto the graph of the function f\left(x\right), then the function f\left(x\right) will clearly approach these lines but never meet them.
One might reasonably assume that:
- \lim\limits_{x \rightarrow \pm \infty} f\left(x\right) = 1 which would indicate that f\left(x\right) seems to have y = 1 as an asymptote.
- \lim\limits_{x \rightarrow -0.5^-} = -\infty and \lim\limits_{x \rightarrow -0.5^+} = \infty which would indicate that f\left(x\right) seems to have x = -0.5 as an asymptote.
f seems to have the asymptotes y = 1 and x = -0.5.
If one overlays the lines y = 2 and x = -4 onto the graph of the function f\left(x\right), then the function f\left(x\right) will clearly approach these lines but never meet them.
One might reasonably assume that:
- \lim\limits_{x \rightarrow \pm \infty} f\left(x\right) = 2 which would indicate that f\left(x\right) seems to have y = 2 as an asymptote.
- \lim\limits_{x \rightarrow -4^-} = \infty and \lim\limits_{x \rightarrow -4^+} = -\infty which would indicate that f\left(x\right) seems to have x = -4 as an asymptote.
f seems to have the asymptotes y = 2 and x = -4.
If one overlays the lines y = 4 onto the graph of the function f\left(x\right), then the function f\left(x\right) will clearly approach this line but never meet it.
One might reasonably assume that \lim\limits_{x \rightarrow -\infty} f\left(x\right) = 4, which would indicate that f\left(x\right) seems to have y = 4 as an asymptote.
f seems to have the asymptote y = 4.
If one overlays the lines y = 1 and y=-3 onto the graph of the function f\left(x\right), then the function f\left(x\right) will clearly approach these lines but never meet them.
One might reasonably assume that:
- \lim\limits_{x \rightarrow \infty} f\left(x\right) = 1 which would indicate that f\left(x\right) seems to have y = 1 as an asymptote.
- \lim\limits_{x \rightarrow -\infty} = -3 which would indicate that f\left(x\right) seems to have y = -3 as an asymptote.
f seems to have the asymptotes y = 1 and y = -3.
If one overlays the lines x = 3 and x = -3 onto the graph of the function f\left(x\right), then the function f\left(x\right) will clearly approach these lines but never meet them.
One might reasonably assume that:
- \lim\limits_{x \rightarrow -3^+} f\left(x\right) = -\infty which would indicate that f\left(x\right) seems to have x=-3 as an asymptote.
- \lim\limits_{x \rightarrow 3^-} = -\infty which would indicate that f\left(x\right) seems to have x = 3 as an asymptote.
f seems to have the asymptotes x = 3 and x = -3.
If one overlays the lines y = -3 onto the graph of the function f\left(x\right), then the function f\left(x\right) will clearly approach this line but never meet it.
One might reasonably assume that:
\lim\limits_{x \rightarrow \pm \infty} f\left(x\right) = -3 which would indicate that f\left(x\right) seems to have y = -3 as an asymptote.
f seems to have the asymptote y = -3.
If one overlays the line y = 5 onto the graph of the function f\left(x\right), then the function f\left(x\right) will clearly approach the line y = 5 but never meet it.
One might reasonably assume that f\left(x\right) will continue to approach the y = 5 as x increases leading to the assumption that \lim\limits_{x \rightarrow \infty} f\left(x\right) = 5.
f seems to have an asymptote at the infinity of equation y=5.