This tutorial provides an analytical approach to rational functions, with the goal of understanding their equations, graphs, and asymptotic behavior. Each example includes a detailed solution. Answers to the matched exercises are available here.
Find the equation of the rational function \( f \) of the form
\[ f(x) = \frac{2}{bx + c} \]whose graph has:
Since the graph has a y-intercept at \( (0,-1) \), we evaluate the function at \( x = 0 \):
\[ f(0) = -1 \]Substituting into the formula gives:
\[ -1 = \frac{2}{c} \]Solving for \( c \):
\[ c = -2 \]A vertical asymptote occurs when the denominator is zero. Since the vertical asymptote is at \( x = 2 \), we set:
\[ 2b + c = 0 \]Substituting \( c = -2 \):
\[ 2b - 2 = 0 \quad \Rightarrow \quad b = 1 \]Therefore, the equation of the rational function is:
\[ f(x) = \frac{2}{x - 2} \]The graph below confirms the y-intercept and vertical asymptote.
Find the equation of the rational function \( f \) of the form
\[ f(x) = \frac{-1}{bx + c} \]whose graph has:
Find the equation of the rational function \( f \) of the form
\[ f(x) = \frac{x + a}{bx + c} \]whose graph has:
The x-intercept occurs where the numerator is zero:
\[ 2 + a = 0 \quad \Rightarrow \quad a = -2 \]The horizontal asymptote is given by the ratio of the leading coefficients of the numerator and denominator:
\[ \frac{1}{b} = \frac{1}{2} \quad \Rightarrow \quad b = 2 \]The vertical asymptote occurs where the denominator is zero. Since it is at \( x = -1 \):
\[ - b + c = 0 \]Substituting \( b = 2 \):
\[ -2 + c = 0 \quad \Rightarrow \quad c = 2 \]Thus, the equation of the rational function is:
\[ f(x) = \frac{x - 2}{2x + 2} \]The graph below confirms the intercepts and asymptotes.
Find the equation of the rational function \( f \) of the form
\[ f(x) = \frac{ax - 2}{bx + c} \]whose graph has:
Find the equation of the rational function \( f \) such that:
A removable discontinuity (hole) at \( x = 1 \) means that both the numerator and denominator contain the factor \( (x - 1) \), which cancels.
A vertical asymptote at \( x = -2 \) means the denominator contains a factor \( (x + 2) \) that does not cancel.
Since the horizontal asymptote is \( y = 1 \), the degrees of the numerator and denominator must be equal, with equal leading coefficients.
A suitable general form is:
\[ f(x) = \frac{(x - 1)(x + a)}{(x - 1)(x + 2)} \]Canceling the common factor gives:
\[ f(x) = \frac{x + a}{x + 2}, \quad x \ne 1 \]Now use the y-intercept condition \( f(0) = -2 \):
\[ \frac{a}{2} = -2 \quad \Rightarrow \quad a = -4 \]Therefore, the rational function is:
\[ f(x) = \frac{(x - 1)(x - 4)}{(x - 1)(x + 2)}, \quad x \ne 1 \]This function has:
Find the equation of a rational function \( f \) such that:
The x-intercepts correspond to the zeros of the numerator. Therefore, the numerator must contain the factors \( (x + 1) \) and \( (x - 2) \):
\[ \text{Numerator} = (x + 1)(x - 2) \]A vertical asymptote at \( x = 1 \) means the denominator contains the factor \( (x - 1) \).
To obtain an oblique asymptote, the degree of the numerator must be exactly one greater than the degree of the denominator.
A suitable general form is:
\[ f(x) = \frac{(x + 1)(x - 2)}{x - 1} \]Now verify the y-intercept by evaluating \( f(0) \):
\[ f(0) = \frac{(1)(-2)}{-1} = -2 \]The y-intercept condition is satisfied.
Therefore, the required rational function is:
\[ f(x) = \frac{(x + 1)(x - 2)}{x - 1} \]This function has: