# Find all the vertical and horizontal asymptotes, maximum and minimum, and points of inflection of...

## Question:

Find all the vertical and horizontal asymptotes, maximum and minimum, and points of inflection of the graph and sketch the graph.

{eq}y = \frac{x+2}{x^2+2x} {/eq}

## Graphing Rational Functions:

Given a rational function {eq}R(x) = \frac{p(x)}{q(x)}{/eq} where {eq}p(x) = a_1x^n + a_2x^{n-1} + \cdot\cdot\cdot {/eq} and {eq}q(x) = b_1x^m + b_2x^{m-1} + \cdot\cdot\cdot{/eq} are polynomial functions of {eq}x{/eq}.

**Vertical Asymptotes**

The vertical asymptotes of {eq}R{/eq} are the zeroes of the denominator {eq}q(x){/eq}. However, if {eq}x=x_0{/eq} is both a zero of {eq}p(x){/eq} and {eq}q(x){/eq}, then {eq}R{/eq} has a hole at {eq}x=x_0{/eq}.

**Horizontal Asymptote**

Comparing the degrees {eq}n {/eq} and {eq}m{/eq} of {eq}p(x){/eq} and {eq}q(x){/eq}, respectively. If:

- {eq}n < m{/eq}, then the x-axis is the horizontal asymptote of the rational function;

- {eq}n=m{/eq}, then the line {eq}y=\frac{a}{b}{/eq} is the horizontal asymptote of the rational function; or

- {eq}n>m{/eq}, then there is no horizontal asymptote but an oblique asymptote to the graph of the rational function.

**Maximum, Minimum and Inflection Points**

The first-order derivative {eq}R'(x){/eq} of the rational function gives the slope of the function as well as the location of the local extremes;

while the second-order derivative {eq}R''(x){/eq} determines its concavity as well as the location of the inflection points and if the local extreme is a local maximum or minimum.

Specifically, given {eq}x = a{/eq}, if:

- {eq}R'(a)=0{/eq} and {eq}R''(a)>0{/eq}, then {eq}x=a{/eq} is a relative minimum;

- {eq}R'(a)=0{/eq} and {eq}R''(a)<0{/eq}, then {eq}x=a{/eq} is a relative maximum;

- {eq}R''(a)=0{/eq}, then {eq}x=a{/eq} is an inflection point.

## Answer and Explanation:

Given the rational function {eq}R(x) = \frac{x+2}{x^2+2x} = \frac{x+2}{x(x+2)}{/eq}.

**Vertical Asymptotes**

Taking the zeroes of the denominator,

{eq}...

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View this answerGiven the rational function {eq}R(x) = \frac{x+2}{x^2+2x} = \frac{x+2}{x(x+2)}{/eq}.

**Vertical Asymptotes**

Taking the zeroes of the denominator,

{eq}\begin{align*} &x(x+2) = 0 \\ &x = 0 \text{ and } x+2 = 0 \\ &x = 0 \text{ and } x=-2 \\ \end{align*} {/eq}

But since {eq}x=-2{/eq} is also a zero of the numerator, there is no vertical asymptote but a hole instead in the rational function at that point.

As such, {eq}\boxed{ VA: \,x=0} {/eq}

Solving for the y-coordinate of the hole, we simplify the rational function, cancelling out common factors in both numerator and denominator, such that:

{eq}R(x) = \frac{1}{x} \\ R(x=-2) = -\frac{1}{2} {/eq}

Therefore, there is a **hole** at the point {eq}\boxed{ \left(-2,-\frac{1}{2} \right)} {/eq}.

**Horizontal Asymptote**

Comparing the degrees of the numerator and denominator, {eq}n=1{/eq} and {eq}m=2{/eq},

since {eq}n<m{/eq}, the horizontal asymptote of the rational function is the {eq}\boxed{x-axis} {/eq}.

**Local Extremes and Inflection Points**

Solving for the first-derivative {eq}R'(x){/eq}:

{eq}\begin{align*} R'(x) &= \frac{d}{dx} \left( \frac{x+2}{x^2+2x} \right) & \text{[Differentiate with respect to } x \text{]} \\ &= \frac{ (x^2+2x)(1) - (x+2)(2x+2)}{(x^2+2x)^2} \\ &= \frac{-(x+2)^2}{x^2(x+2)^2} \\ &= -\frac{1}{x^2} \end{align*} {/eq}