SOLUTION: Please help me with these problem did any of this before. Give an example of each of the following: a. A function with vertical asymptotes at x=-3 and x=1, and a horizont

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Please help me with these problem did any of this before. Give an example of each of the following: a. A function with vertical asymptotes at x=-3 and x=1, and a horizont      Log On


   

Question 1027807: Please help me with these problem did any of this before.
Give an example of each of the following:
a. A function with vertical asymptotes at x=-3 and x=1, and a horizontal asymptote at y=2.
b. A word problem whose solution is 12!/9!3!
c. The equation of an ellipse with vertices at (0,0) and (-8,0).
d. A logarithmic expression equivalent to 5.

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
a. To have those vertical asymptotes, a rational functions should have in the denominator the factors
%28x-%28-3%29%29=%28x%2B3%29 and %28x-1%29 .
With %28x%2B3%29%28x-1%29=x%5E2%2B2x-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%5E2 is a suitable numerator for that, and its value for x=-3 and for x=1 is not zero.
So, highlight%28f%28x%29=2x%5E2%2F%28x%5E2%2B2x-3%29%29 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%28x%29=%28x%2B2%29%2F%28x-1%29 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%2B2=3 ,
but as x approaches 1, the denominator approaches zero,
making abs%28f%28x%29%29 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%28x%29=%28x-1%29%28x%2B2%29%2F%28x-1%29 does not exist for x=1 , but f%28x%29=x%2B2 for all x%3C%3E1 , so within its domain f%28x%29 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%28x%29=%28x-1%29%28x%2B2%29%2F%28x-1%29%5E2 is {{g(x)=(x+2)/(x-1)}}} for x%3C%3E1 ,
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%28x%29=%282x%5E2%2B1%29%2Fx%5E2=2%2B1%2Fx%5E2 has y=2 for a vertical asymptote,
but so does .

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 .
But there are 3%21=3%2A2%2A1 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%21%2F9%21 lists of 3 members, but since each set of 3 can be listed 3%21 ways, there are really 12%21%2F%289%21%2A3%21%29 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
%28x%2B4%29%5E2%2F16%2By%5E2=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-%28-8%29=8 , the semi-major axis is 8%2F2=4 , and the equation of the ellipse must be %28x%2B4%29%5E2%2F16%2By%5E2%2Fb%5E2=1 with any b (the semi-minor axis) such that 0%3Cb%3C4 .

d. log%28100000%29 is a logarithmic expression equivalent to 5. So is log%282%2C32%29 .
EXPLANATION:
100000=10%5E5 so log%28100000%29=log%2810%2C100000%29=log%2810%2C10%5E5%29=5 , and
2%5E5=32 so log%282%2C32%29=log%282%2C2%5E5%29=5 .