# How do you graph f(x)=-3/(x-1) using holes, vertical and horizontal asymptotes, x and y intercepts?

Feb 6, 2018

Holes: $\textcolor{g r e e n}{\text{Since nothing was cancelled out, there are no holes.}}$

Vertical Asymptote: Use the denominator. The function is undefined when the denominator is zero. At what value of $x$ will the denominator equal zero?

$\textcolor{g r e e n}{\text{Set the denominator to zero, and then solve for x}}$.

$\textcolor{g r e e n}{\text{x - 1 = 0}}$
$\textcolor{g r e e n}{\text{x = 1}}$

$\textcolor{g r e e n}{\text{The vertical asymptote will be at x = 1}}$.

Horizontal Asymptote: Use the degrees of the numerator and denominator.

• If the degree of the numerator is less than the degree of the denominator, HA is at y=0.
• If the degree of the numerator is more than the degree of the denominator, there is no HA.
• If they are equal, divide the coefficient of ${x}^{\mathrm{de} g r e e}$ in the numerator by the coefficient of ${x}^{\mathrm{de} g r e e}$ in the denominator.

$\textcolor{g r e e n}{\text{The degree of the numerator is zero. The degree of the denominator is 1. The HA is at y=0.}}$

x-int: The graph will pass through the x-axis when y is equal to zero.

$\textcolor{g r e e n}{\text{Set y = 0, then solve for x.}}$

$\textcolor{g r e e n}{0} \textcolor{g r e e n}{=} - \frac{\textcolor{g r e e n}{3}}{\textcolor{g r e e n}{x - 1}}$

$\textcolor{g r e e n}{\text{In this case, this is not possible, therefore there is no x-intercept.}}$

y-int: The graph will pass through the y-axis when x is equal to zero.

$\textcolor{g r e e n}{\text{Set x = 0, then solve for y.}}$

$\textcolor{g r e e n}{y} \textcolor{g r e e n}{=} - \frac{\textcolor{g r e e n}{3}}{\textcolor{g r e e n}{0 - 1}}$

$\textcolor{g r e e n}{y} \textcolor{g r e e n}{=} \textcolor{g r e e n}{3}$

$\textcolor{g r e e n}{\text{The graph passes through the y-axis at 3.}}$

Use these values to plot points, then use a table of values if extra points are needed.