Question

graph the following function on the given coordinate plane and identify the indicated attributes.
$f(x)=\frac{x^{2}-3x-10}{x^{2}-4}$
identify the vertical asymptotes:
$x=2$
identify the horizontal asymptotes:
$y=1$
identify the domain:
$(-\infty,2)\cup(2,\infty)$
identify the range:
$(-\infty,1)\cup(1,\infty)$

a rational function is shown below:
$f(x)=\frac{x^{2}+x-6}{3x^{2}-12}$
determine the asymptotic restrictions on the domain: (vertical asymptotes)
$x=-2 \quad x=2$
determine the domain:
$(-\infty,-2)\cup(-2,2)\cup(2,\infty)$
determine the range:
$(-\infty, \frac{1}{3})\cup(\frac{1}{3}, \frac{5}{12})\cup(\frac{5}{12},\infty)$

Explanation:

(First Function: $f(x)=\frac{x^2-3x-10}{x^2-4}$):

Step1: Factor numerator and denominator

Numerator: $x^2-3x-10=(x-5)(x+2)$
Denominator: $x^2-4=(x-2)(x+2)$
Simplify: $f(x)=\frac{(x-5)(x+2)}{(x-2)(x+2)}=\frac{x-5}{x-2}$ (for $x
eq-2$)

Step2: Find vertical asymptotes

Set simplified denominator to 0: $x-2=0 \implies x=2$
(Note: $x=-2$ is a hole, not an asymptote)

Step3: Find horizontal asymptotes

Compare leading terms: $\frac{x^2}{x^2}=1 \implies y=1$

Step4: Find domain

Exclude $x$ that make original denominator 0: $x
eq-2, 2$
Domain: $(-\infty,-2)\cup(-2,2)\cup(2,\infty)$

Step5: Find range

Rewrite $y=\frac{x-5}{x-2}$: $y(x-2)=x-5 \implies xy-2y=x-5 \implies x(y-1)=2y-5 \implies x=\frac{2y-5}{y-1}$
Exclude $y=1$, and check hole at $x=-2$: $y=\frac{-2-5}{-2-2}=\frac{7}{4}$
Range: $(-\infty,1)\cup(1,\infty)$ (since $\frac{7}{4}$ is within this interval)

(Second Function: $f(x)=\frac{x^2+x-6}{3x^2-12}$):

Step1: Factor numerator and denominator

Numerator: $x^2+x-6=(x+3)(x-2)$
Denominator: $3x^2-12=3(x^2-4)=3(x-2)(x+2)$
Simplify: $f(x)=\frac{(x+3)(x-2)}{3(x-2)(x+2)}=\frac{x+3}{3(x+2)}$ (for $x
eq2$)

Step2: Find vertical asymptotes

Set simplified denominator to 0: $3(x+2)=0 \implies x=-2$
(Note: $x=2$ is a hole, not an asymptote)

Step3: Find horizontal asymptotes

Compare leading terms: $\frac{x^2}{3x^2}=\frac{1}{3} \implies y=\frac{1}{3}$

Step4: Find domain

Exclude $x$ that make original denominator 0: $x
eq-2, 2$
Domain: $(-\infty,-2)\cup(-2,2)\cup(2,\infty)$

Step5: Find range

Rewrite $y=\frac{x+3}{3(x+2)}$: $3y(x+2)=x+3 \implies 3xy+6y=x+3 \implies x(3y-1)=3-6y \implies x=\frac{3-6y}{3y-1}$
Exclude $y=\frac{1}{3}$, and check hole at $x=2$: $y=\frac{2+3}{3(2+2)}=\frac{5}{12}$
Range: $(-\infty,\frac{1}{3})\cup(\frac{1}{3},\frac{5}{12})\cup(\frac{5}{12},\infty)$

Corrected Answers:

For $f(x)=\frac{x^2-3x-10}{x^2-4}$:

• Vertical Asymptotes: $x=2$
• Horizontal Asymptotes: $y=1$
• Domain: $(-\infty,-2)\cup(-2,2)\cup(2,\infty)$
• Range: $(-\infty,1)\cup(1,\infty)$

For $f(x)=\frac{x^2+x-6}{3x^2-12}$:

• Vertical Asymptotes: $x=-2$
• Horizontal Asymptotes: $y=\frac{1}{3}$
• Domain: $(-\infty,-2)\cup(-2,2)\cup(2,\infty)$
• Range: $(-\infty,\frac{1}{3})\cup(\frac{1}{3},\frac{5}{12})\cup(\frac{5}{12},\infty)$

Answer:

(Second Function: $f(x)=\frac{x^2+x-6}{3x^2-12}$):

Step1: Factor numerator and denominator

Numerator: $x^2+x-6=(x+3)(x-2)$
Denominator: $3x^2-12=3(x^2-4)=3(x-2)(x+2)$
Simplify: $f(x)=\frac{(x+3)(x-2)}{3(x-2)(x+2)}=\frac{x+3}{3(x+2)}$ (for $x
eq2$)

Step2: Find vertical asymptotes

Set simplified denominator to 0: $3(x+2)=0 \implies x=-2$
(Note: $x=2$ is a hole, not an asymptote)

Step3: Find horizontal asymptotes

Compare leading terms: $\frac{x^2}{3x^2}=\frac{1}{3} \implies y=\frac{1}{3}$

Step4: Find domain

Exclude $x$ that make original denominator 0: $x
eq-2, 2$
Domain: $(-\infty,-2)\cup(-2,2)\cup(2,\infty)$

Step5: Find range

Rewrite $y=\frac{x+3}{3(x+2)}$: $3y(x+2)=x+3 \implies 3xy+6y=x+3 \implies x(3y-1)=3-6y \implies x=\frac{3-6y}{3y-1}$
Exclude $y=\frac{1}{3}$, and check hole at $x=2$: $y=\frac{2+3}{3(2+2)}=\frac{5}{12}$
Range: $(-\infty,\frac{1}{3})\cup(\frac{1}{3},\frac{5}{12})\cup(\frac{5}{12},\infty)$

Corrected Answers:

For $f(x)=\frac{x^2-3x-10}{x^2-4}$:

• Vertical Asymptotes: $x=2$
• Horizontal Asymptotes: $y=1$
• Domain: $(-\infty,-2)\cup(-2,2)\cup(2,\infty)$
• Range: $(-\infty,1)\cup(1,\infty)$

For $f(x)=\frac{x^2+x-6}{3x^2-12}$:

• Vertical Asymptotes: $x=-2$
• Horizontal Asymptotes: $y=\frac{1}{3}$
• Domain: $(-\infty,-2)\cup(-2,2)\cup(2,\infty)$
• Range: $(-\infty,\frac{1}{3})\cup(\frac{1}{3},\frac{5}{12})\cup(\frac{5}{12},\infty)$