Answer:
• The slant asymptote is y=x-3.
,
• The function g(x) has a removable discontinuity at x=-1.
,
• The function g(x) has a non-removable discontinuity at x=2.
Explanation:
Given that the function, f(x) has zeros of 8, -1, and -3.
Then, by the factor theorem:
[tex]f(x)=(x-8)(x+1)(x+3)[/tex]
If we divide the f(x) by x²-x-2:
[tex]\begin{gathered} g(x)=\frac{f(x)}{x^2-x-2}=\frac{(x-8)(x+1)(x+3)}{x^2-2x+x-2} \\ =\frac{(x-8)(x+1)(x+3)}{x(x-2)+1(x-2)} \\ =\frac{(x-8)(x+1)(x+3)}{(x+1)(x-2)} \\ =\frac{(x-8)(x+3)}{x-2} \end{gathered}[/tex]
Therefore, the rational function, g(x) in simplest form is:
[tex]g(x)=\frac{(x-8)(x+3)}{x-2}[/tex]
Since the degree of the numerator is one degree greater than the degree of the denominator, a slant asymptote exists. To find this, divide the numerator by the denominator:
[tex]g(x)=\frac{x^2+3x-8x-24}{x-2}=\frac{x^2-5x-24}{x-2}[/tex]
The quotient of the division is x-3 with a remainder of 18.
The slant asymptote is y=x-3.
Given g(x) in the form below:
[tex]g(x)=\frac{(x-8)(x+1)(x+3)}{(x+1)(x-2)}[/tex]
Set the denominator equal to 0 to find the point of discontinuities.
[tex]\begin{gathered} (x+1)(x-2)=0 \\ x+1=0\text{ or }x-2=0 \\ x=-1,x=2 \end{gathered}[/tex]
The function is discontinuous at x=-1 and x=2.
However, notice that there is a common factor in the numerator and the denominator, x+1. The zero for this factor is x = -1 . Therefore:
• The function g(x) has a removable discontinuity at x=-1.
• The function g(x) has a non-removable discontinuity at x=2.
