\(\text{www.the}\beta\text{etafunction.com}\)
Home
Login
Questions: Algebra BusinessCalculus
Please login to create an exam or a quiz.
Find the horizontal and vertical asymptote(s) of \(\displaystyle f(x) = \frac{x^{2} + 9 x - 10}{8 x^{2} + 24 x - 80}\), if they exist.
The degree of the numerator is the same as the degree of the denominator, so the horizontal asymptote is the ratio of the leading coefficients \(\displaystyle y=\frac{1}{8}\). The vertical asymptotes are the zeros of the denominator. \(\displaystyle 8 x^{2} + 24 x - 80=0 \iff 8 \left(x - 2\right) \left(x + 5\right)=0\). The vertical asymptotes are \(\displaystyle x=-5\) and \(\displaystyle x=2\).
\begin{question}Find the horizontal and vertical asymptote(s) of $f(x) = \frac{x^{2} + 9 x - 10}{8 x^{2} + 24 x - 80}$, if they exist.
\soln{9cm}{The degree of the numerator is the same as the degree of the denominator, so the horizontal asymptote is the ratio of the leading coefficients $y=\frac{1}{8}$. The vertical asymptotes are the zeros of the denominator. $8 x^{2} + 24 x - 80=0 \iff 8 \left(x - 2\right) \left(x + 5\right)=0$. The vertical asymptotes are $x=-5$ and $x=2$.}
\end{question}
\documentclass{article}
\usepackage{tikz}
\usepackage{amsmath}
\usepackage[margin=2cm]{geometry}
\usepackage{tcolorbox}
\newcounter{ExamNumber}
\newcounter{questioncount}
\stepcounter{questioncount}
\newenvironment{question}{{\noindent\bfseries Question \arabic{questioncount}.}}{\stepcounter{questioncount}}
\renewcommand{\labelenumi}{{\bfseries (\alph{enumi})}}
\newif\ifShowSolution
\newcommand{\soln}[2]{%
\ifShowSolution%
\noindent\begin{tcolorbox}[colframe=blue,title=Solution]#2\end{tcolorbox}\else%
\vspace{#1}%
\fi%
}%
\newcommand{\hideifShowSolution}[1]{%
\ifShowSolution%
%
\else%
#1%
\fi%
}%
\everymath{\displaystyle}
\ShowSolutiontrue
\begin{document}\begin{question}(10pts) The question goes here!
\soln{9cm}{The solution goes here.}
\end{question}\end{document}<p> <p>Find the horizontal and vertical asymptote(s) of <img class="equation_image" title=" \displaystyle f(x) = \frac{x^{2} + 9 x - 10}{8 x^{2} + 24 x - 80} " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%20%3D%20%5Cfrac%7Bx%5E%7B2%7D%20%2B%209%20x%20-%2010%7D%7B8%20x%5E%7B2%7D%20%2B%2024%20x%20-%2080%7D%20" alt="LaTeX: \displaystyle f(x) = \frac{x^{2} + 9 x - 10}{8 x^{2} + 24 x - 80} " data-equation-content=" \displaystyle f(x) = \frac{x^{2} + 9 x - 10}{8 x^{2} + 24 x - 80} " /> , if they exist.</p> </p><p> <p>The degree of the numerator is the same as the degree of the denominator, so the horizontal asymptote is the ratio of the leading coefficients <img class="equation_image" title=" \displaystyle y=\frac{1}{8} " src="/equation_images/%20%5Cdisplaystyle%20y%3D%5Cfrac%7B1%7D%7B8%7D%20" alt="LaTeX: \displaystyle y=\frac{1}{8} " data-equation-content=" \displaystyle y=\frac{1}{8} " /> . The vertical asymptotes are the zeros of the denominator. <img class="equation_image" title=" \displaystyle 8 x^{2} + 24 x - 80=0 \iff 8 \left(x - 2\right) \left(x + 5\right)=0 " src="/equation_images/%20%5Cdisplaystyle%208%20x%5E%7B2%7D%20%2B%2024%20x%20-%2080%3D0%20%5Ciff%208%20%5Cleft%28x%20-%202%5Cright%29%20%5Cleft%28x%20%2B%205%5Cright%29%3D0%20" alt="LaTeX: \displaystyle 8 x^{2} + 24 x - 80=0 \iff 8 \left(x - 2\right) \left(x + 5\right)=0 " data-equation-content=" \displaystyle 8 x^{2} + 24 x - 80=0 \iff 8 \left(x - 2\right) \left(x + 5\right)=0 " /> . The vertical asymptotes are <img class="equation_image" title=" \displaystyle x=-5 " src="/equation_images/%20%5Cdisplaystyle%20x%3D-5%20" alt="LaTeX: \displaystyle x=-5 " data-equation-content=" \displaystyle x=-5 " /> and <img class="equation_image" title=" \displaystyle x=2 " src="/equation_images/%20%5Cdisplaystyle%20x%3D2%20" alt="LaTeX: \displaystyle x=2 " data-equation-content=" \displaystyle x=2 " /> .</p> </p>