where will the discontinuities occur in the graph of the rational function?\nf(x)=\frac{x^{2}-4x}{x^{2}-x…

where will the discontinuities occur in the graph of the rational function?\nf(x)=\frac{x^{2}-4x}{x^{2}-x - 12}\na. at (x = 4)\nb. at (x=-3)\nc. at (x = 4) and (x=-3)\nd. at (x = 0,x = 4) and (x=-3)

where will the discontinuities occur in the graph of the rational function?\nf(x)=\frac{x^{2}-4x}{x^{2}-x - 12}\na. at (x = 4)\nb. at (x=-3)\nc. at (x = 4) and (x=-3)\nd. at (x = 0,x = 4) and (x=-3)

Answer

Explanation:

Step1: Factor the numerator and denominator

The numerator $x^{2}-4x=x(x - 4)$. The denominator $x^{2}-x - 12=(x - 4)(x+3)$. So $f(x)=\frac{x(x - 4)}{(x - 4)(x + 3)}$.

Step2: Find the values that make the denominator zero

Set the denominator $(x - 4)(x + 3)=0$. Using the zero - product property, we have $x-4 = 0$ or $x+3=0$. Solving $x-4 = 0$ gives $x = 4$, and solving $x+3=0$ gives $x=-3$. When $x = 4$, the common factor $(x - 4)$ in the numerator and denominator can be canceled out (it is a removable discontinuity), but when $x=-3$, the function is undefined.

Answer:

B. at $x=-3$