# Analyze the function f(x)= (x^3-4x^2-31x+70)/(x^2-5x+6). Find the y-intercept, the...

## Question:

Analyze the function

{eq}f(x)= \frac{x^3 - 4x^2 - 31x + 70}{x^2 - 5x + 6} {/eq}.

Find the y-intercept, the x-intercept(s), the removable singularities, the vertical asymptotes and the horizontal asymptotes.

1) How many x-intercepts does {eq}f {/eq} have?

2) How many removable singularities does {eq}f {/eq} have?

3) How many vertical asymptotes does {eq}f {/eq} have?

4) How many horizontal asymptotes does {eq}f {/eq} have?

What is the y-intercept of the graph {eq}y=f(x) {/eq}?

What are the x-intercepts of the graph {eq}y=f(x) {/eq}?

What is the x-value of the removable singularity?

What is the limit of {eq}f(x) {/eq} as {eq}x {/eq} approaches this removable singularity?

What is the equation of the vertical asymptote?

What is the limit of {eq}f(x) {/eq} as {eq}x {/eq} approaches this vertical asymptote from the positive side?

What is the limit of {eq}f(x) {/eq} as {eq}x {/eq} approaches this vertical asymptote from the negative side?

## Rational Functions

Rational functions are functions which are ratios of polynomials. Rational functions often have vertical asymptotes, horizontal asymptotes, holes (or removable singularities), and slant asymptotes. To determine if a rational function has a vertical asymptote or a hole, the numerator and denominator should be fully factored. Any factors which cancel completely out of the denominator correspond to a removable singularity (or hole) in the graph. Any factors which are still in the denominator correspond to a vertical asymptote. Horizontal asymptotes can be found by comparing the degrees of the numerator and denominator. A higher degree in the numerator results in no horizontal asymptotes, a higher degree in the denominator results in a horizontal asymptote at the x axis, and equal degrees results in a horizontal asymptote at the ratio of the leading coefficients.

## Answer and Explanation:

1) How many x-intercepts does {eq}f {/eq} have?

First notice that the domain of the function is {eq}(\infty, 2)\cup(2,3)\cup(3,\infty) {/eq}. To find x intercepts, set the function equal to zero.

{eq}\frac{x^3 - 4x^2 - 31x + 70}{x^2 - 5x + 6} = 0\\ x^3 - 4x^2 - 31x + 70=0\\ (x-7)(x-2)(x+5)=0\\ x=7,2,-5 {/eq}

But since 2 is not in the domain, we only have two x intercepts {eq}(7,0) , (-5,0) {/eq}

2) How many removable singularities does {eq}f {/eq} have?

The numerator and denominator have one factor in common, and so there is one removable singularity, or hole, at {eq}(2, 35) {/eq}

3) How many vertical asymptotes does {eq}f {/eq} have?

There is one vertical asymptote corresponding to the factor {eq}x-3 {/eq} in the denominator. The vertical asymptote is {eq}x=3 {/eq}

4) How many horizontal asymptotes does {eq}f {/eq} have?

There are no horizontal asymptotes, since the degree of the numerator is larger than the degree of the denominator

What is the y-intercept of the graph {eq}y=f(x) {/eq}?

To find a y intercept, substitute x=0. The y intercept is {eq}(0,\frac{35}{3}) {/eq}

What are the x-intercepts of the graph {eq}y=f(x) {/eq}?

The x intercepts are {eq}(7,0) , (-5,0) {/eq}

What is the x-value of the removable singularity?

The x value of the removable singularity is x=2

What is the limit of {eq}f(x) {/eq} as {eq}x {/eq} approaches this removable singularity?

As x approaches the removable singularity, the function approaches 35.

What is the equation of the vertical asymptote?

{eq}x=3 {/eq}

What is the limit of {eq}f(x) {/eq} as {eq}x {/eq} approaches this vertical asymptote from the positive side?

The limit is {eq}-\infty {/eq}, as the denominator approaches 0 but is positive and the numerator is negative

What is the limit of {eq}f(x) {/eq} as {eq}x {/eq} approaches this vertical asymptote from the negative side?

The limit is {eq}\infty {/eq}, since the denominator approaches 0 but is negative and the numerator is negative

#### Learn more about this topic:

from GMAT Prep: Help and Review

Chapter 10 / Lesson 11