I really need help:(

Question

For the function y = ((x - 1)(x + 2)) / (x - 1), what is true?
 

There is a hole at x = 1, since x=1 is a zero of the numerator and the denominator.

There is a vertical asymptote at x = -2 and a hole at x=1

There are no discontinuities, since x - 1 cancels out.

The is a vertical asymptote at x=1.
 

For the function y = (3x - 1) / (5x + 2), what is true?
 

The vertical asymptote is 3/5.

There is a hole at x = -2/5.

The horizontal asymptote is 0.

The horizontal asymptote is 3/5.
 

For the function y = (x - 1) / ((x + 2)(x - 3)), what is true?
 

There are vertical asymptotes at x = -2 and x = 3 and a hole at x = 1.

There are vertical asymptotes at x = -2 and x = 3 and a horizontal asymptote at x = 1.

There are vertical asymptotes at x = -2 and x = 3.

The horizontal asymptote is y = 1.

Answer 1

First, we need to find some pieces of information about this function.  We know that it is a ration function because it is in fraction form.  So we will have intercepts, asymptotes, and holes.

 

To find the x-intercept, we set the numerator equal to zero and solve for x.

 

(x - 1)(x + 2) = 0

 

x = 1          and            x = -2

 

 

To find the y-intercept, we evaluate f(0).

 

y-intercept is (0, 2)

 

 

To find the vertical asymptote, set the denominator equal to zero and solve for x.

 

x - 1 = 0

 

x = 1

 

 

Since x=1 is a solution and at the same time an undefined value, x=1 is a hole.

 

 

If we simplify the function, we get

 

f(x) = x + 2

 

This is a linear function.   This means we have no vertical or horizontal asymptotes.

 

 

To find possible horizontal asymptotes, we examine the degrees of the numerator and denominator:

 

If degree of numerator is less than the degree of denominator, then horizontal asymptote is   y=0.

If degree of numerator is equal to the degree of denominator, then horizontal asymptote is ratio of the leading terms.

If degree of numerator is greater than the degree of denominator, then we have a slant asymptote.

 

 

So lets review.

 

x-intercept is (-2, 0)

y-intercept is (0, 2)

Hole at (1, 3)

 

 

Use this information to answer your first question.

 

Apply these procedures to answer the next two questions.

Answer 2

For the function y = ((x - 1)(x + 2)) / (x - 1), what is true?

There is a hole at x = 1, since x=1 is a zero of the numerator and the denominator. correct

There is a vertical asymptote at x = -2 and a hole at x=1
There are no discontinuities, since x - 1 cancels out.
The is a vertical asymptote at x=1.

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For the function y = (3x - 1) / (5x + 2), what is true?

The vertical asymptote is 3/5.
There is a hole at x = -2/5. 
The horizontal asymptote is 0.

The horizontal asymptote is 3/5.  correct

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For the function y = (x - 1) / ((x + 2)(x - 3)), what is true?

There are vertical asymptotes at x = -2 and x = 3 and a hole at x = 1.
There are vertical asymptotes at x = -2 and x = 3 and a horizontal asymptote at x = 1.
There are vertical asymptotes at x = -2 and x = 3.  correct
The horizontal asymptote is y = 1.