Question

1. State the domain of the rational function. (2 points)
f(x) = thirteen divided by quantity ten minus x.


a) All real numbers except -10 and 10
b) All real numbers except 13
c) All real numbers except 10
d) All real numbers except -13 and 13

2. State the vertical asymptote of the rational function. (2 points)
f(x) = quantity x minus six times quantity x plus six divided by quantity x squared minus nine.


a) x = 6, x = -6
b) x = 3, x = -3
c) x = -6, x = 6
d) None

3. State the horizontal asymptote of the rational function. (2 points)
f(x) = quantity x squared plus four x minus seven divided by quantity x minus seven.


a) None
b) y = -4
c) y = 7
d) y = 6

4. State the horizontal asymptote of the rational function. (2 points)
f(x) = quantity x squared plus eight x minus two divided by quantity x minus two.


a) None
b) y = 1
c) y = -8
d) y = 2

5. Give an example of a rational function that has a horizontal asymptote at y = 0 and a vertical asymptote at
x = 2 and x = 1.

Answer 1

1)      The domain is every value of x for which f(x) is a real number.

f(x) = 13 / (10-x)
The only x value that would not produce a real number for f(x) is 10, since you cannot divide a number by zero. Answer is C

2)      F(x) =(x-6)(x+6)/(x2  - 9)
The vertical asymptotes are x=3 and x=-3. Graph the function on a graphing calculator to observe the behavior of the function at these points. There is both a positive and negative vertical asymptote a both x=3 and x=-3. Keep in mind that the denominator approaches zero at these points, and thus f(x) approaches either positive or negative infinite, depending on whether the denominator, however small, is a positive or negative number. Answer is B) 3, -3

3)      F(x) = (x2 + 4x-7) / (x-7)
Although there is a vertical asymptote as x=7, there is no horizontal asymptote. This makes sense. As X gets bigger, there is nothing to hold y back from getting greater and greater. X2 is the dominant term, and it’s only in the numerator. A) none

4)      (x2 + 8x -2) / (x-2)
This function is very similar in structure to the previous one. Same rules apply. Dominant term only in the numerator means no horizontal asymptote. A)None

5)      Our function approaches 0 as x approaches infinite, and has a vertical asymptote at x=2 and x=1.
Here’s an easy example: 10 / ((x-2)*(x-1)). At x=2 and x=1, there is both a positive and negative vertical asymptote. As x approaches infinite, the numerator is dominated by the denominator, which contains x (actually x2 ), and thus y approaches zero.  

Answer 2

1. Considering f(x) = $\frac{13}{10-x}$
c) All real numbers except 10
Compare the denominator to zero and it will give you the point where there is no f(x) because it is impossible to divide anything by 0.
10-x=0 ⇔ x=10
The function has a domain containing all real numbers except 10 → x<10 or x<10.

2. Considering f(x)=$\frac{(x-6)(x+6)}{x^{2}-9 }$
b) x = 3, x = -3
Once more, compare the denominator to zero and it will give you the point/points where there is no f(x) because it is impossible to divide anything by 0, and there is where you can find the vertical asymptotes. 
x²-9=0 ⇔ x=-3, x=3
The vertical asymptotes are x=3 and x=-3.

3. Considering $\frac{x^{2}+4x-7}{x-7}$
a) None (of those presented)
Apply long division on the function. 
The horizontal asymptote is y=x+11.

4. Considering $\frac{x^{2}+8x-2}{x-2}$
a) None (of those presented)
Once more, apply long division on the function.
The horizontal asymptote is y=x+10.

5. The polynomial in the numerator has to have a smaller degree than the polynomial in the denominator. The polynomial in the denominator needs to have factors of x-1 and x-2. This factors should not be part of the polynomial in the numerator. An example of such a function could be: f(x)=$\frac{8x-2}{ x^{2} -3x+2}$