Answer:
[tex]f(x)=\dfrac{x^2(x+2)(x-5)}{(x+3)(x-3)(x-5)}[/tex]
Step-by-step explanation:
A rational function is a mathematical expression representing the quotient of two polynomials:
[tex]f(x)=\dfrac{g(x)}{h(x)}[/tex]
In a rational function, vertical asymptotes occur at the values of x for which the denominator equals zero.
Given that the rational function has vertical asymptotes at x = 3 and x = -3, it means that its denominator has zeros at these values. Therefore, two factors of the denominator are (x - 3) and (x + 3).
[tex]f(x)=\dfrac{g(x)}{(x-3)(x+3)}[/tex]
A removable discontinuity, commonly known as a "hole" in a graph, occurs in a rational function when there is a common factor with an x in both the numerator and the denominator, leading to the cancellation of that factor.
Given that the rational function has a removable discontinuity at x = 5, there is the common factor (x - 5) in both the numerator and the denominator of the rational function:
[tex]f(x)=\dfrac{(x-5)}{(x-3)(x+3)(x-5)}[/tex]
The equation of the slant asymptote is the quotient of the division of the numerator of the function by its denominator (ignoring the remainder). Given that the slant asymptote is x + 2, the quotient of the numerator divided by denominator must be (x + 2), so:
[tex]f(x)=\dfrac{(x+2)(x-5)}{(x+3)(x-3)(x-5)}[/tex]
Finally, a slant asymptote occurs in a rational function when the degree of the numerator polynomial is exactly one greater than the degree of the denominator polynomial. Since the denominator of the rational function is degree 3, the numerator needs to be degree 4, so we can simply add the factor of x² to the numerator to make it degree 4:
[tex]f(x)=\dfrac{x^2(x+2)(x-5)}{(x+3)(x-3)(x-5)}[/tex]
This function has:
- Vertical asymptotes at x = 3 and x = -3.
- Slant asymptote at x + 2.
- Removable discontinuity at x = 5.
The graph of the function is attached. Note that the asymptotes are shown as dashed red lines, and the removable discontinuity is marked by an open circle.
