# Identify the extent, leading coefficient, and constant coefficient in the following polynomial...

## Question:

Identify the extent, leading coefficient, and constant coefficient in the following polynomial functions:

(a) {eq}f(x) = 9x^2 - 6x + 16 {/eq}

(b) {eq}f(x) = - \frac{2}{3}x^3 - 5x + 3x^2 + 10 {/eq}

(c) {eq}f(x) = 3 - 6x {/eq}

Determine the vertical and horizontal asymptotes and the domain of the following rational functions:

(d) {eq}f(x) = \frac{6}{x - 10} {/eq}

(e) {eq}g(x) = \frac{2x - 3}{x - 8} {/eq}

(f) {eq}h(x) = \frac{9}{x} {/eq}

## Polynomial and Rational Functions

A polynomial function of degree {eq}n {/eq} can be written in the form {eq}f(x) = a_0 + a_1x + a_2x^2 + \cdots + a_nx^n {/eq} where {eq}a_0 {/eq} is the constant term and {eq}a_n {/eq} is the leading coefficient.

A rational function can be written in the form {eq}\displaystyle Q(x) = \frac{n(x)}{d(x)} {/eq} where {eq}n(x) {/eq} and {eq}d(x) {/eq} are polynomials.

To find the domain of {eq}Q(x) {/eq}, solve {eq}d(x) =0 {/eq} and exclude any solutions from set {eq}\mathbb{R} {/eq}.

To find the vertical asymptotes of {eq}Q(x) {/eq}, find all values of {eq}x {/eq} for which {eq}d(x) = 0 {/eq} but {eq}n(x) \neq 0 {/eq}.

If the degree of {eq}n(x) {/eq} is less than the degree of {eq}d(x) {/eq}, then {eq}y = 0 {/eq} is a horizontal asymptote of {eq}Q(x) {/eq}.

If the degree of {eq}n(x) {/eq} is equal to the degree of {eq}d(x) {/eq}, then {eq}\displaystyle y = \frac{a_n}{b_n} {/eq} is a horizontal asymptote of {eq}Q(x) {/eq} where {eq}a_n {/eq} is the leading coefficient of {eq}d(x) {/eq} and {eq}b_n {/eq} is the leading coefficient of {eq}n(x) {/eq}.

## Answer and Explanation:

**Question 1**

(a) The degree of the polynomial {eq}f(x) = 9x^2 - 6x + 16 {/eq} is {eq}2 {/eq}. The leading coefficient is {eq}9 {/eq}. The constant term is {eq}16 {/eq}.

(b) The degree of the polynomial {eq}\displaystyle f(x) = - \frac{2}{3}x^3 - 5x + 3x^2 + 10 {/eq} is {eq}3 {/eq}. The leading coefficient is {eq}- \frac{2}{3} {/eq}. The constant term is {eq}10 {/eq}.

(c) The degree of the polynomial {eq}\displaystyle f(x) = f(x) = 3 - 6x {/eq} is {eq}1 {/eq}. The leading coefficient is {eq}-6 {/eq}. The constant term is {eq}3 {/eq}.

**Question 2**

(d) Solve {eq}x-10 = 0 {/eq} to obtain {eq}x = 10 {/eq}. The domain of {eq}\displaystyle f(x) = \frac{6}{x - 10} {/eq} is {eq}(-\infty,10)\cup(10,\infty) {/eq}. The vertical asymptote is {eq}x = 10 {/eq}. The horizontal asymptote is {eq}y = 0 {/eq} since the degree of the numerator is less than the degree of the denominator.

(e) Solve {eq}x-8= 0 {/eq} to obtain {eq}x = 8 {/eq}. The domain of {eq}\displaystyle g(x) = \frac{2x - 3}{x - 8} {/eq} is {eq}(-\infty,8)\cup(8,\infty) {/eq}. The vertical asymptote is {eq}x = 8 {/eq}. The horizontal asymptote is {eq}y = \frac{2}{1} = 2 {/eq} since the degree of the numerator is equal to the degree of the denominator.

(f) Note that the denominator is zero when {eq}x = 0 {/eq}. The domain of {eq}\displaystyle h(x) = \frac{9}{x} {/eq} is {eq}(-\infty,0)\cup(0,\infty) {/eq}. The vertical asymptote is {eq}x = 0 {/eq}. The horizontal asymptote is {eq}y = 0 {/eq} since the degree of the numerator is less than the degree of the denominator.

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Chapter 10 / Lesson 11