Question

members of group:
destiny valori

1. for ( f(x) = \frac{6x + 12}{2x - 4} )

a. what type of function is this? how do you know?

b. what is the domain?

c. write the equation of any vertical asymptote(s) of ( f ).

d. find any horizontal asymptote of ( f ) and write it as an equation.

e. find any vertical intercept(s) of ( f ), and write it as ordered pair(s).

f. find any horizontal intercept(s) of ( f ), and write as ordered pair(s).

g. sketch the function. be sure to label the asymptotes using dotted lines and label intercepts as ordered pairs.

graph with x-axis from -10 to 10 and y-axis from -10 to 10, grid lines

Explanation:

Step1: Identify the function type

A rational function is of the form \( f(x)=\frac{p(x)}{q(x)} \) where \( p(x) \) and \( q(x) \) are polynomials and \( q(x)
eq0 \). Here \( p(x) = 6x + 12 \) and \( q(x)=2x - 4 \), so \( f(x) \) is a rational function.

Step2: Find the domain

The denominator \( q(x)=2x - 4\). Set \( 2x-4
eq0\). Solving \( 2x-4 = 0\) gives \( x = 2\). The domain is all real numbers except \( x = 2\), written as \( (-\infty,2)\cup(2,\infty) \).

Step3: Find the vertical asymptote

The vertical asymptote occurs at the value of \( x \) that makes the denominator zero (after simplifying, if possible). Since \( f(x)=\frac{6x + 12}{2x - 4}=\frac{6(x + 2)}{2(x - 2)}=\frac{3(x + 2)}{x - 2}\) (no common factors in numerator and denominator other than constants), the vertical asymptote is \( x = 2 \).

Step4: Find the horizontal asymptote

For a rational function \( f(x)=\frac{a_nx^n+\cdots}{b_mx^m+\cdots}\), if \( n=m \), the horizontal asymptote is \( y=\frac{a_n}{b_m}\). Here \( n = m=1\), \( a_1 = 6\), \( b_1=2\), so \( y=\frac{6}{2}=3 \).

Step5: Find the vertical intercept

Set \( x = 0\). \( f(0)=\frac{6(0)+12}{2(0)-4}=\frac{12}{-4}=-3 \). The vertical - intercept is the ordered pair \( (0,-3) \).

Step6: Find the horizontal intercept

Set \( y = f(x)=0\). Then \( \frac{6x + 12}{2x - 4}=0\), which implies \( 6x+12 = 0\) (since a fraction is zero when the numerator is zero and the denominator is non - zero). Solving \( 6x+12 = 0\) gives \( x=-2 \). The horizontal - intercept is the ordered pair \( (-2,0) \).

Answer:

a. Rational function. It is in the form \( f(x)=\frac{p(x)}{q(x)} \) where \( p(x) \) and \( q(x) \) are polynomials.
b. \( (-\infty,2)\cup(2,\infty) \)
c. \( x = 2 \)
d. \( y = 3 \)
e. \( (0,-3) \)
f. \( (-2,0) \)
g. Sketch: Draw the vertical asymptote \( x = 2 \) (dotted line), horizontal asymptote \( y = 3 \) (dotted line), plot the intercepts \( (0,-3) \) and \( (-2,0) \). For \( x\lt2\) (e.g., \( x = 1\), \( f(1)=\frac{6(1)+12}{2(1)-4}=\frac{18}{-2}=-9\)), for \( x\gt2\) (e.g., \( x = 3\), \( f(3)=\frac{6(3)+12}{2(3)-4}=\frac{30}{2}=15\)). Then sketch the two branches of the rational function approaching the asymptotes.