# Graph f(x) = \frac{5x}{x^2 - 25} and indicate: a) the domain b) the intercepts and asymptotes c)...

## Question:

Graph {eq}f(x) = \frac{5x}{x^2 - 25} {/eq} and indicate:

a) the domain

b) the intercepts and asymptotes

c) where the function is increasing and decreasing

d) intervals of local maximum and local minimum

e) where the function is concave upward and concave downward

f) any inflection points

## Rational Functions:

Given a rational function {eq}F(x) = \frac{P(x)}{Q(x)} {/eq}

The domain of {eq}F(x){/eq} is the set of all real numbers {eq}R{/eq} except

(i) the zeroes of {eq}Q(x){/eq}, and

(ii) set of {eq}x{/eq} that will make the radicand of an even-based radical expression negative.

The zeroes of {eq}Q(x){/eq} are the vertical asymptotes of {eq}F(x){/eq}.

{eq}{/eq}

Horizontal Asymptotes

Let {eq}F(x) = \frac{ax^n + ...}{bx^m + ...} {/eq}.

(i) If {eq}n<m{/eq}, then {eq}F(x){/eq} has a horizontal asymptote at the {eq}x{/eq}-axis;

(ii) If {eq}n=m{/eq}, then {eq}F(x){/eq} has a horizontal asymptote at {eq}y=\frac{a}{b}{/eq}; and,

(iii) If {eq}n>m{/eq}, then {eq}F(x){/eq} has no horizontal asymptote but an oblique asymptote.

{eq}{/eq}

The {eq}x{/eq}-intercept/s of {eq}F(x){/eq} are the zeroes of {eq}P(x){/eq} in the domain of {eq}F(x){/eq}; while

the {eq}y{/eq}-intercept of {eq}F(x){/eq} is equal to {eq}F(x=0{/eq} if {eq}x=0{/eq} is in the domain of {eq}F(x){/eq}.

{eq}{/eq}

Using the first-order derivative of {eq}F(x){/eq}, denoted as {eq}F'(x){/eq}, the function is:

(i) increasing if {eq}F'(x)>0{/eq}; and

(ii) decreasing if {eq}F'(x)<0{/eq}.

The zeroes of {eq}F'(x){/eq} also gives the relative/local extrema of {eq}F(x){/eq}.

Using {eq}F''(x){/eq}, the second-order derivative of {eq}F(x){/eq}, determines if the relative extrema is a:

(i) relative maximum if {eq}F''(x)<0{/eq}; and,

(ii) relative minimum if {eq}F''(x)>0{/eq}.

{eq}{/eq}

Concavity

{eq}F(x){/eq}

(i) is concave up if {eq}F''(x)>0{/eq};

(ii) is concave down if {eq}F''(x)<0{/eq}; and,

(iii) has an inflection point if {eq}F''(x)=0{/eq}.

Given {eq}f(x)=\frac{5x}{x^2 - 25}{/eq}.

a) The domain is the set of all real numbers {eq}R{/eq} except the zeroes of {eq}x^2-25{/eq}.

{eq}x^2 - 25...

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