How do you find the vertical, horizontal or slant asymptotes for #h(x)=5^(x-2)#?

Question

Ariana Alvarez · Accepted Answer

Horizontal : #y = 0 larr#

graph{y(y-5^x/25)=0 [-1.95, 1.948, -0.975, 0.975]}

#h=5^(x-2)=5^x5^(-2)=1/25 5^x>=0#

y-intercept ( x = 0 ) : 1/25

As #x to oo, y to oo#.

As #x to -oo, y to 0#.

So, the x-axis y = 0 is the asymptote.

Ariana Alvarez · Answer

undefined To find the vertical, horizontal, or slant asymptotes for the function $ h(x) = 5^{x-2} $, follow these guidelines:

1. Vertical asymptotes: There are no vertical asymptotes for exponential functions.

2. Horizontal asymptote: To find the horizontal asymptote, consider the behavior of the function as $ x $ approaches positive or negative infinity.

$$ \lim_{x \to \pm\infty} 5^{x-2} = 5^{-2} = \frac{1}{25} $$

So, the horizontal asymptote is $ y = \frac{1}{25} $.

3. Slant asymptote: Exponential functions do not have slant asymptotes.

In summary, for the function $ h(x) = 5^{x-2} $:
- There are no vertical asymptotes.
- The horizontal asymptote is $ y = \frac{1}{25} $.
- There are no slant asymptotes.