To find the vertical and horizontal asymptotes of the rational function [tex]\( f(x) = \frac{5 - 9x}{5 + 2x} \)[/tex], we need to follow these steps:
### Vertical Asymptotes
Vertical asymptotes occur where the denominator of the function is equal to zero, as the function will tend toward infinity or negative infinity at these points.
1. Set the denominator equal to zero:
[tex]\[
5 + 2x = 0
\][/tex]
2. Solve for [tex]\( x \)[/tex]:
[tex]\[
2x = -5 \implies x = -\frac{5}{2}
\][/tex]
So, the vertical asymptote is at [tex]\( x = -\frac{5}{2} \)[/tex].
### Horizontal Asymptotes
Horizontal asymptotes are determined by analyzing the end behavior of the function as [tex]\( x \)[/tex] approaches infinity or negative infinity. Specifically, we need to find the limits of the function as [tex]\( x \to \infty \)[/tex] and [tex]\( x \to -\infty \)[/tex].
1. To find the limit as [tex]\( x \to \infty \)[/tex]:
[tex]\[
\lim_{{x \to \infty}} \frac{5 - 9x}{5 + 2x}
\][/tex]
2. Considering the degrees of the numerator and the denominator are equal, the horizontal asymptote will be the ratio of the leading coefficients:
[tex]\[
\lim_{{x \to \infty}} \frac{-9x}{2x} = \frac{-9}{2}
\][/tex]
3. Similarly, to find the limit as [tex]\( x \to -\infty \)[/tex]:
[tex]\[
\lim_{{x \to -\infty}} \frac{5 - 9x}{5 + 2x}
\][/tex]
4. Again, the horizontal asymptote will be the same as the ratio of the leading coefficients:
[tex]\[
\lim_{{x \to -\infty}} \frac{-9x}{2x} = \frac{-9}{2}
\][/tex]
So, the horizontal asymptote is [tex]\( y = -\frac{9}{2} \)[/tex].
### Summary
- The vertical asymptote is [tex]\( x = -\frac{5}{2} \)[/tex].
- The horizontal asymptote is [tex]\( y = -\frac{9}{2} \)[/tex].
These vertical and horizontal asymptotes describe where the graph of the rational function [tex]\( f(x) = \frac{5 - 9x}{5 + 2x} \)[/tex] approaches infinity and how it behaves at extreme values of [tex]\( x \)[/tex].
