Question

worksheet b: (topics 1.9 - 1.10) ap style mcq - va and holes
name:
graph of f
directions: the rational function f is shown above and has a vertical asymptote at x = 4, a hole at x = 6, and a horizontal asymptote of y = 5. use the graph of f to answer the following questions.

1. which of the following limit statements about f is correct?

(a) lim f(x)=4 (b) lim f(x)=4 (c) lim f(x)=∞ (d) lim f(x)=∞

1. which of the following statements about f is correct?

(a) lim f(x)=6 (b) lim f(x)=6 (c) lim f(x)=5 (d) lim f(x)=4

1. which of the following limit statements about f is correct?

(a) lim f(x)= -∞ (b) lim f(x)=∞ (c) lim f(x)=5 (d) lim f(x)=4

1. which of the following could be the equation for f(x)?

(a) f(x)=-\frac{(5x - 22)(x - 6)}{(x - 6)(x - 4)}
(b) f(x)=\frac{(5x - 22)(x - 6)}{(x - 6)(x - 4)}
(c) f(x)=-\frac{(5x - 22)(x - 4)}{(x - 6)(x - 4)}
(d) f(x)=-\frac{(x - 6)(x - 4)}{(5x - 22)(x - 6)}

Explanation:

Step1: Analyze vertical asymptote

A vertical asymptote at $x = 4$ means the denominator of the rational - function has a factor of $(x - 4)$ and the function approaches $\pm\infty$ as $x$ approaches 4. Also, a hole at $x = 6$ means both the numerator and denominator have a factor of $(x - 6)$.

Step2: Analyze horizontal asymptote

A horizontal asymptote of $y = 5$ means the ratio of the leading - coefficients of the numerator and denominator (when the degrees of the numerator and denominator are the same) is 5.

Question 1

The function has a vertical asymptote at $x = 4$. As $x$ approaches 4 from the left or right, the function approaches $\pm\infty$. Since the graph goes up as $x$ approaches 4 from the right, $\lim_{x
ightarrow4^{+}}f(x)=\infty$. So the answer to question 1 is (D).

Question 2

Since there is a hole at $x = 6$, the limit as $x$ approaches 6 exists. The horizontal asymptote is $y = 5$, so $\lim_{x
ightarrow6}f(x)=5$. The answer to question 2 is (C).

Question 3

As $x$ approaches $\pm\infty$, the function approaches the horizontal asymptote $y = 5$. So $\lim_{x
ightarrow\pm\infty}f(x)=5$. The answer to question 3 is (C).

Question 4

We know the function has a factor of $(x - 6)$ in both the numerator and denominator (for the hole at $x = 6$) and a factor of $(x - 4)$ in the denominator (for the vertical asymptote at $x = 4$). Also, since the horizontal asymptote is $y = 5$, the ratio of the leading - coefficients of the numerator and denominator (when the degrees are the same) should be 5.
For option (A), $f(x)=-\frac{(5x - 22)(x - 6)}{(x - 6)(x - 4)}=-\frac{5x^{2}-52x + 132}{x^{2}-10x + 24}$, the ratio of the leading - coefficients is - 5.
For option (B), $f(x)=\frac{(5x - 22)(x - 6)}{(x - 6)(x - 4)}=\frac{5x^{2}-52x + 132}{x^{2}-10x + 24}$, the ratio of the leading - coefficients is 5.
For option (C), $f(x)=-\frac{(5x - 22)(x - 4)}{(x - 6)(x - 4)}=-\frac{5x^{2}-42x + 88}{x^{2}-10x + 24}$, the ratio of the leading - coefficients is - 5.
For option (D), $f(x)=-\frac{(x - 6)(x - 4)}{(5x - 22)(x - 6)}=-\frac{x^{2}-10x + 24}{5x^{2}-52x + 132}$, the ratio of the leading - coefficients is $-\frac{1}{5}$.
So the answer to question 4 is (B).

Answer:

1. D. $\lim_{x

ightarrow4^{+}}f(x)=\infty$

1. C. $\lim_{x

ightarrow6}f(x)=5$

1. C. $\lim_{x

ightarrow\pm\infty}f(x)=5$

1. B. $f(x)=\frac{(5x - 22)(x - 6)}{(x - 6)(x - 4)}$