What is a vertical asymptote in a rational function?
When does a vertical asymptote occur at x = a in a rational function?
State what happens to the values of a rational function r near a vertical asymptote x = a.
A polynomial function has the following real zeros:
x=-2 (in the numerator and denominator)
x=3 (only in the denominator)
Based on this information, does the graph of this polynomial function have a vertical asymptote? If so, at which x-value?
Consider a rational function where x=5 is a real zero in the denominator and not in the numerator. What does this imply about the graph of the rational function at x=5?
Determine where the vertical asymptote of the following rational functions. If there isn't any, explain why.
r \left( x \right) = \dfrac{x^2 - 5x + 6}{x - 2}
r \left( x \right) = \dfrac{x^3 - 3x^2 + 3x - 1}{x - 1}
r \left( x \right) = \dfrac{x^3 - 4x^2 + 6x - 4}{x - 2}
Determine the values of x of the following rational functions for which r \left( x \right) is undefined:
r \left( x \right) = \dfrac{x^2 - 9}{x^2 - 4}
r \left( x \right) = \dfrac{x^3 - 8}{x - 2}
r \left( x \right) = \dfrac{x^3 - 4x^2 + 6x - 4}{x - 2}
r \left( x \right) = \dfrac{x^3 - 3x^2 + 3x - 1}{x - 1}
r \left( x \right) = \dfrac{x^4 - 16x^2 + 64}{x - 4}
Determine the limit of the following rational functions as x approaches the given value. If the limit does not exist, explain why.
\lim_{x \to 1} \dfrac{x^3 - 1}{x^2 - 1}
\lim_{x \to 0} \dfrac{x^2 - 4x}{x^2 + x - 6}
\lim_{x \to -2} \dfrac{x^4 - 16}{x^3 + 8}
\lim_{x \to 3} \dfrac{x^3 - 27}{x^2 - 9}
For the rational function h\left(x\right) = \dfrac{x^2 + 2x - 3}{x^2 - 4}, explain the role of real zeros in the denominator and numerator in determining the vertical asymptotes.
Predict the behavior of the rational function g(x) = \dfrac{3x^2 + 5x - 2}{x^2 - 9} near the vertical asymptotes at x = -3 and x = 3.
Consider the following rational function: f(x) = \dfrac{x^3 - 2x^2 + x}{x^2 - x}.
Determine the multiplicity of the real zero:
Numerator
Denominator
For each of the following rational function graphs, identify the vertical asymptote(s).
f(x)=\dfrac{x^2-1}{x^2-x-6}
f(x)=\dfrac{2x}{x^2-4}
f(x)=\dfrac{x^2+4x+4}{x^2-9}
f(x)=\dfrac{x^3-8}{x^2-4x+4}
Given the graphs of the following rational functions, compare their behavior near the vertical asymptotes:
r(x)=\dfrac{x^2-4}{x^2-1}
s(x)=\dfrac{x^3-27}{x^2-9}
For the rational function r \left( x \right) = \dfrac{x^4 - 16x^2 + 64}{x - 4}:
Explain why there is a vertical asymptote at x = 4.
How would this change if the numerator was x^4 - 16x^2 + 63 instead?
Consider the two rational functions p(x) = \dfrac{x^2 - x - 12}{x^3 - 3x^2 - x + 3} and q(x) = \dfrac{x - 3}{x^2 + 2x - 15}.
Identify the vertical asymptotes of each function.
Compare the behavior of the functions near their vertical asymptotes and explain the differences.
Given the rational function r \left( x \right) = \dfrac{x^3 + x^2 - 4x - 4}{x - 2}:
Explain the behavior of the function as x approaches 2 from the left and from the right.
How does this relate to the vertical asymptote?
Analyze the behavior of the rational function g(x) = \dfrac{2x^2 - 5x - 3}{x^3 - 6x^2 + 9x} near its vertical asymptotes:
Rewrite the given rational function in a form that clearly shows the vertical asymptotes.
Determine the vertical asymptotes and predict the behavior of the function near each asymptote.
Explain the effect of the multiplicity of the real zeros on the vertical asymptotes.