Question

homework assignment #6
category: quiz
current learning objective: identifying vertical asymptotes of rational functions
question 23 practice similar questions
score: 0 of 1 point
let ( f(x)=\frac{3 x^{2}-11 x - 20}{3 x^{2}+8 x + 5} )
find the vertical asymptotes
a ( x=\frac{5}{3}, x = 1 )
b ( x=-\frac{5}{3}, x = 1 )
c ( x=-\frac{5}{3}, x=-1 )
d ( x=\frac{5}{3}, x=-1 )
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Explanation:

Step1: Factor numerator and denominator

For the numerator \(3x^{2}-11x - 20\), using the formula \(ax^{2}+bx + c\), we find two numbers \(m\) and \(n\) such that \(m\times n=3\times(- 20)=-60\) and \(m + n=-11\). We get \(m=-15\) and \(n = 4\).

$$3x^{2}-11x - 20=3x^{2}-15x+4x - 20=3x(x - 5)+4(x - 5)=(3x + 4)(x-5)$$

For the denominator \(3x^{2}+8x + 5\), we find two numbers \(p\) and \(q\) such that \(p\times q=3\times5 = 15\) and \(p + q=8\). We get \(p = 3\) and \(q = 5\).

$$3x^{2}+8x + 5=3x^{2}+3x+5x + 5=3x(x + 1)+5(x + 1)=(3x + 5)(x + 1)$$

So \(f(x)=\frac{(3x + 4)(x - 5)}{(3x + 5)(x + 1)}\)

Step2: Find vertical asymptotes

Vertical asymptotes occur at the values of \(x\) that make the denominator equal to zero (while the numerator is not zero at those values).
Set the denominator \((3x + 5)(x + 1)=0\)
\(3x+5 = 0\) gives \(x=-\frac{5}{3}\)
\(x + 1=0\) gives \(x=-1\)
We check that the numerator is not zero at \(x =-\frac{5}{3}\) and \(x=-1\).
When \(x=-\frac{5}{3}\), \((3\times(-\frac{5}{3})+4)(-\frac{5}{3}-5)=(-5 + 4)(-\frac{5}{3}-5)
eq0\)
When \(x=-1\), \((3\times(-1)+4)(-1 - 5)=(1)(-6)
eq0\)

Answer:

C. \(x =-\frac{5}{3},x=-1\)