Question 1027807
a. To have those vertical asymptotes, a rational functions should have in the denominator the factors
{{{(x-(-3))=(x+3)}}} and {{{(x-1)}}} .
With {{{(x+3)(x-1)=x^2+2x-3}}} for a denominator
a rational function with {{{y=2}}} for a horizontal asymptote
must have for a numerator a quadratic polynomial with {{{2}}} for a leading coefficient.
{{{2x^2}}} is a suitable numerator for that, and its value for {{{x=-3}}} and for {{{x=1}}} is not zero.
So, {{{highlight(f(x)=2x^2/(x^2+2x-3))}}} has all the required asymptotes.

FURTHER EXPLANATION:
A rational function may have vertical asymptotes at the values of {{{x}}} that make
the denominator zero.
So, {{{f(x)=(x+2)/(x-1)}}} does not exist for {{{x=1}}} ,
and has {{{x=1}}} as a vertical asymptote,
because at valued close to {{{x=1}}} the numerator is close to {{{1+2=3}}} ,
but as {{{x}}} approaches {{{1}}}, the denominator approaches zero,
making {{{abs(f(x))}}} increase without bounds.
However, if a value of {{{x}}} makes both, numerator and denominator zero,
there may not be a vertical asymptote at that value.
For example, {{{f(x)=(x-1)(x+2)/(x-1)}}} does not exist for {{{x=1}}} , but {{{f(x)=x+2}}} for all {{{x<>1}}} , so within its domain {{{f(x)}}} is just a linear function,
and graphs as a slanted line, with a hole at the point (1,3), and no vertical asymptote.
On the other hand, {{{g(x)=(x-1)(x+2)/(x-1)^2}}} is {{g(x)=(x+2)/(x-1)}}} for {{{x<>1}}} ,
and has {{{x=1}}} for a vertical asymptote.
A rational function has a horizontal asymptote {{{y=k}}} for some number {{{k}}}
when numerator and denominator are polynomials of the same degree, and the ratio of the leading coefficients is {{{k}}}.
It is obvious that {{{f(x)=(2x^2+1)/x^2=2+1/x^2}}} has {{{y=2}}} for a vertical asymptote,
but so does {{{g(x)=(2x^5+x^4+x^3+x^2+x+1)/(x^5-x^4+x^3-x^2+x-1)}}} .
 
b. How many different possibilities has an academic coach trying to choose a team of 3 students to represent the school in an academic contest? (The order the team members are chosen or listed does not matter).
EXPLANATION:
Choosing a set of {{{3}}} from a larger set of {{{12}}} , is a question of combinations, because the choosing order, position, or role of each team member is not important.
When choosing, there are {{{12}}} possibilities for the first choice, {{{12-1=11}}} for the second choice, and {{{1-2=10}}} for the third choice.
That is {{{12*11*10=12*11*10*9*8*7*6*5*4*3*2*1/(9*8*7*6*5*4*3*2*1)=12!/9!}}} .
But there are {{{3!=3*2*1}}} ways to choose the same team, because any of the {{[3}}} members could have been chosen first, followed in each case, by any of the {{[3-1=2}}} other members, leaving just {{{2-1=1}}} choice.
So there may be {{{12!/9!}}} lists of {{{3}}} members, but since each set of {{{3}}} can be listed {{{3!}}} ways, there are really {{{12!/(9!*3!)}}} teams.
NOTE: An alternate word problem is
How many different cheerleading team of {{{9}}} members can be chosen from {{{12}}} students showing up at the tryouts?
 
c. The equation of an ellipse with vertices at (0,0) and (-8,0) could be
{{{(x+4)^2/16+y^2=1}}} .
EXPLANATION:
The points called vertices are the ends of the major axis, since the ends of the minor axis are usually called co-vertices.
The given points are on a horizontal line (axis), so the segments between them has to be the horizontal major axis.
Since the distance between the vertices is {{{0-(-8)=8}}} , the semi-major axis is {{{8/2=4}}} , and the equation of the ellipse must be {{{(x+4)^2/16+y^2/b^2=1}}} with any {{{b}}} (the semi-minor axis) such that {{{0<b<4}}} .
 
d. {{{log(100000)}}} is a logarithmic expression equivalent to 5. So is {{{log(2,32)}}} .
EXPLANATION:
{{{100000=10^5}}} so {{{log(100000)=log(10,100000)=log(10,10^5)=5}}} , and
{{{2^5=32}}} so {{{log(2,32)=log(2,2^5)=5}}} .