Question

directions: for each of the following

1. ( f ( x ) = \frac { ( x + 2 ) ( x - 6 ) } { x - 3 } )

left:
( x = 2.9 )
( lim _ { x
ightarrow 3 ^ { - } } f ( x ) = infty )
right:
( x = 3.1 )
( lim _ { x
ightarrow 3 ^ { + } } f ( x ) = - infty )
directions: write an equation of a ratio
10 the graph of ( f ) has a hole at ( x = 3 )
1.9 - 1.10 rational functions vertical asymptotes and hol

Explanation:

Step1: Analyze the function

For a rational function \( f(x)=\frac{(x + 2)(x - 6)}{x - 3}\), a vertical asymptote occurs when the denominator is zero and the numerator is non - zero at that point.
When \(x = 3\), the denominator \(x-3=0\), and the numerator \((3 + 2)(3 - 6)=5\times(-3)=-15
eq0\).

Step2: Calculate the left - hand limit

Let \(x = 3-\epsilon\) where \(\epsilon>0\) and \(\epsilon\to0\).
\(f(x)=\frac{(x + 2)(x - 6)}{x - 3}=\frac{( (3-\epsilon)+ 2)((3-\epsilon)- 6)}{(3-\epsilon)- 3}=\frac{(5-\epsilon)(-3-\epsilon)}{-\epsilon}\)
As \(\epsilon\to0\), the numerator \((5-\epsilon)(-3-\epsilon)\approx5\times(-3)=-15\) (for small \(\epsilon\)), and the denominator \(-\epsilon\to0^{-}\). So \(\lim_{x\to3^{-}}f(x)=\infty\)

Step3: Calculate the right - hand limit

Let \(x = 3+\epsilon\) where \(\epsilon>0\) and \(\epsilon\to0\).
\(f(x)=\frac{(x + 2)(x - 6)}{x - 3}=\frac{( (3+\epsilon)+ 2)((3+\epsilon)- 6)}{(3+\epsilon)- 3}=\frac{(5+\epsilon)(-3+\epsilon)}{\epsilon}\)
As \(\epsilon\to0\), the numerator \((5+\epsilon)(-3+\epsilon)\approx5\times(-3)=-15\) (for small \(\epsilon\)), and the denominator \(\epsilon\to0^{+}\). So \(\lim_{x\to3^{+}}f(x)=-\infty\)

Answer:

The vertical asymptote is \(x = 3\), \(\lim_{x\to3^{-}}f(x)=\infty\), \(\lim_{x\to3^{+}}f(x)=-\infty\)