Given: (x^3 + x^2 - 22x - 40)/(x^4 - x^3 - 7x^2 + x + 6)
This type of function is called a rational function.
(N(x))/(D(x)) = (a_nx^n + ...)/(b_b x^m + ...), where a_n " & " b_m are the leading coefficients and n and m are the degrees of the two functions.
color(blue)"Factor" both the numerator and the denominator need to be factored.
The easiest way is to graph each function individually and find the zeros (x-intercepts).
This is the graph of N(x) = x^3 + x^2 - 22x - 40
graph{x^3 + x^2 - 22x - 40 [-5, 6, -80, 10]}
It has x-intercepts at x = -4, -2 and 5
This means color (red)(N(x)) = x^3 + x^2 - 22x - 40 = color (red)((x +4)(x+2)(x-5))
This is the graph of D(x) = x^4 - x^3 - 7x^2 + x + 6:
graph{x^4 - x^3 - 7x^2 + x + 6 [-5, 6, -15, 10]}
It has x-intercepts at x = -2, -1, 1 and 3
This means color (red)(D(x)) = x^4 - x^3 - 7x^2 + x + 6 = color(red)((x+2)(x+1)(x-1)(x-3))
color(blue)"Factored function:"
f(x) = ((x +4)cancel((x+2))(x-5))/(cancel((x+2))(x+1)(x-1)(x-3))
color(magenta)("Finding the Domain depends on holes, and vertical asymptotes.")
If you can cancel a factor that is both in the numerator and the denominator, that is where you have a hole; a removable discontinuity.
color (magenta)("Hole at " x = -2
Vertical asymptotes are found by setting D(x) = 0 after the holes are eliminated.
color (magenta)("Vertical asymptotes are at " x = -1, 1, x = 3
Domain: (-oo, -2) uu (-2, -1) uu (-1, 1) uu (1, 3) uu (3, oo)
