a) The polynomial in expanded form is [tex]f(x) = x^{3}-2\cdot x^{2}-21\cdot x -18[/tex].
b) The slant asymptote is represented by the linear function is [tex]y = -x + 1[/tex].
c) There is a discontinuity at [tex]x = 2[/tex] with a slant asymptote.
a) In this question we are going to use the Factor Theorem, which establishes that polynomial are the result of products of binomials of the form [tex]x-r_{i}[/tex], where [tex]r_{i}[/tex] is the i-th root of the polynomial and the grade is equal to the quantity of roots. Therefore, the polynomial [tex]f(x)[/tex] has the following form:
[tex]f(x) = (x-6)\cdot (x+1)\cdot (x+3)[/tex]
And the expanded form is obtained by some algebraic handling:
[tex]f(x) = (x-6)\cdot (x^{2}+4\cdot x +3)[/tex]
[tex]f(x) = x\cdot (x^{2}+4\cdot x + 3)-6\cdot (x^{2}+4\cdot x +3)[/tex]
[tex]f(x) = x^{3}+4\cdot x^{2}+3\cdot x -6\cdot x^{2}-24\cdot x -18[/tex]
[tex]f(x) = x^{3}-2\cdot x^{2}-21\cdot x -18[/tex] (1)
The polynomial in expanded form is [tex]f(x) = x^{3}-2\cdot x^{2}-21\cdot x -18[/tex].
b) In this question we divide the polynomial found in a) (in factor form) by the polynomial [tex]x^{2}-x -2[/tex] (also in factor form). That is:
[tex]g(x) = \frac{(x-6)\cdot (x+1)\cdot (x+3)}{(x-2)\cdot (x+1)}[/tex]
[tex]g(x) = \frac{(x-6)\cdot (x+3)}{x-2}[/tex] (2)
The slant asymptote is defined by linear function, whose slope ([tex]m[/tex]) and intercept ([tex]b[/tex]) are determined by the following expressions:
[tex]m = \lim_{x \to \pm \infty} \frac{g(x)}{x}[/tex] (3)
[tex]b = \lim_{x \to \pm \infty} [g(x)-x][/tex] (4)
If [tex]g(x) = \frac{(x-6)\cdot (x+3)}{x-2}[/tex], then the equation of the slant asymptote is:
[tex]m = \lim_{x \to 2} \frac{(x-6)\cdot (x+3)}{x\cdot (x-2)}[/tex]
[tex]m = \lim_{x \to \pm \infty} \left(\frac{x^{2}-3\cdot x -18}{x^{2}-2\cdot x} \right)[/tex]
[tex]m = 1[/tex]
[tex]b = \lim_{x \to \pm \infty} \left(\frac{x^{2}-3\cdot x -18}{x-2}-x \right)[/tex]
[tex]b = \lim_{x \to \pm \infty} \left(\frac{x^{2}-3\cdot x - 18-x^{2}+2\cdot x}{x-2}\right)[/tex]
[tex]b = \lim_{n \to \infty} \left(\frac{-x-18}{x-2} \right)[/tex]
[tex]b = -1[/tex]
The slant asymptote is represented by the linear function is [tex]y = -x + 1[/tex].
c) The number of discontinuities in rational functions is equal to the number of binomials in the denominator, which was determined in b). Hence, we have a discontinuity at [tex]x = 2[/tex] with a slant asymptote.
We kindly invite to check this question on asymptotes: https://brainly.com/question/4084552
