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

iescabroussexg

iescabroussexg

Answered question

2022-09-12

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

Answer & Explanation

Karma Estes

Karma Estes

Beginner2022-09-13Added 11 answers

Holes: 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?

Set the denominator to zero, and then solve for x.

x - 1 = 0
x = 1

The vertical asymptote will be at x = 1.

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

1. If the degree of the numerator is less than the degree of the denominator, HA is at y=0.
2. If the degree of the numerator is more than the degree of the denominator, there is no HA.
3. If they are equal, divide the coefficient of x d e g r e e in the numerator by the coefficient of x d e g r e e in the denominator.

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.

Set y = 0, then solve for x.

0 = - 3 x - 1

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.

Set x = 0, then solve for y.

y = - 3 0 - 1

y=3

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.

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