Question

complete the following for the function ( f(x) = \frac{4x^3 + 3x^2 + 3x + 3}{x^2 + 4} ).

a. find the slant asymptote of ( f ).

b. find the vertical asymptote(s) of ( f ) (if any).

c. graph ( f ) and all of its asymptotes with a graphing utility.

a. the slant asymptote is ( y = square ).

Explanation:

Part a: Slant Asymptote

Step1: Perform Polynomial Long Division

Divide the numerator \(4x^3 + 3x^2 + 3x + 3\) by the denominator \(x^2 + 4\).

• Divide \(4x^3\) by \(x^2\) to get \(4x\). Multiply \(x^2 + 4\) by \(4x\) to get \(4x^3 + 16x\).
• Subtract this from the numerator: \((4x^3 + 3x^2 + 3x + 3)-(4x^3 + 16x)=3x^2 - 13x + 3\).
• Now divide \(3x^2\) by \(x^2\) to get \(3\). Multiply \(x^2 + 4\) by \(3\) to get \(3x^2 + 12\).
• Subtract this: \((3x^2 - 13x + 3)-(3x^2 + 12)= -13x - 9\).

So, \(\frac{4x^3 + 3x^2 + 3x + 3}{x^2 + 4}=4x + 3+\frac{-13x - 9}{x^2 + 4}\).

Step2: Determine Slant Asymptote

As \(x\to\pm\infty\), the remainder term \(\frac{-13x - 9}{x^2 + 4}\) approaches \(0\) (since the degree of the numerator is less than the degree of the denominator). So the slant asymptote is the linear part of the quotient, which is \(y = 4x + 3\).

Part b: Vertical Asymptote(s)

Step1: Analyze Denominator

To find vertical asymptotes, we set the denominator equal to zero and solve for \(x\): \(x^2 + 4 = 0\).

Step2: Solve the Equation

\(x^2=-4\), which has no real solutions (since the square of a real number cannot be negative). So there are no vertical asymptotes.

Part c: Graphing

(This part is done using a graphing utility. You would plot the function \(f(x)=\frac{4x^3 + 3x^2 + 3x + 3}{x^2 + 4}\), the slant asymptote \(y = 4x + 3\), and note there are no vertical asymptotes. The graph of the function will approach the slant asymptote as \(x\to\pm\infty\) and have no vertical breaks since the denominator is never zero for real \(x\).)

Answer:

s:
a. The slant asymptote is \(y=\boxed{4x + 3}\)
b. There are no vertical asymptotes.
c. (Graphing done via utility, with \(f(x)\), \(y = 4x + 3\) plotted, no vertical asymptotes.)