Question

question 32, 5.4.37
part 10 of 13
follow the steps for graphing a rational function to graph the function ( r(x)=\frac{x^{2}+14 x + 48}{x + 8} )
the function has no vertical asymptote.
determine the horizontal asymptote(s), if any exist. select the correct choice and, if necessary, fill in the answer box(es) within your choic
a. the function has one horizontal asymptote,
(type an equation. use integers or fractions for any numbers in the equation.)
b. the function has two horizontal asymptotes. the top asymptote is and the bottom asymptote is
(type equations. use integers or fractions for any numbers in the equations.)
c. the function has no horizontal asymptote.
determine the oblique asymptote(s), if any exist. select the correct choice and, if necessary, fill in the answer box(es) within your choic
a. the function has one oblique asymptote,
(type an equation. use integers or fractions for any numbers in the equation.)
b. the function has two oblique asymptotes. the oblique asymptote with a negative slope is and the oblique asymptote with a positive
slope is
(type equations. use integers or fractions for any numbers in the equations.)
c. the function has no oblique asymptote.

Explanation:

Step1: Simplify the rational function

Factor the numerator \(x^{2}+14x + 48=(x + 6)(x+8)\). Then \(R(x)=\frac{(x + 6)(x + 8)}{x + 8}=x+6\) for \(x
eq - 8\).

Step2: Analyze horizontal asymptote

For a rational function \(y=\frac{f(x)}{g(x)}\), if \(\text{deg}(f(x))-\text{deg}(g(x))=1\) (after simplification, here \(f(x)=x + 6\) (degree \(1\)) and \(g(x)=1\) (degree \(0\))), the rule for horizontal asymptotes (which is based on the ratio of the leading - coefficients when \(\text{deg}(f(x))=\text{deg}(g(x))\)) does not apply.

Step3: Analyze oblique asymptote

Since \(R(x)=x + 6\) (a linear function) for \(x
eq-8\), and an oblique asymptote is a non - horizontal, non - vertical linear asymptote. But when we simplify the rational function \(R(x)=\frac{x^{2}+14x + 48}{x + 8}\) (after canceling the common factor \(x + 8\) for \(x
eq-8\)), the function \(y=x + 6\) is a straight line. However, the concept of an asymptote is for a function that approaches a line as \(x\to\pm\infty\) in the 'true' rational - function sense (without simplification that removes a factor). But if we consider the original form \(\frac{x^{2}+14x + 48}{x + 8}\), we can also use polynomial long division. Dividing \(x^{2}+14x + 48\) by \(x + 8\): \(x^{2}+14x+48=(x + 8)(x+6)+0\).

Answer:

C. The function has no horizontal asymptote.
A. The function has one oblique asymptote, \(y=x + 6\)