Question

Identify the hole(removable discontinuity) of the function F(x) =x^2-2x-24/x+4 give your answer as an ordered pair.

Answer 1

Answer:

$\lparen x,y)=\left(-4,0\right)$

Explanation:

We were given the function:

$f\left(x\right)=\frac{x^2-2x-24}{x+4}$

Discontinuity occurs when the denominator equals zero

$\begin{gathered} x+4=0 \\ x=-4 \end{gathered}$

When x = -4, the value of the denominator is zero

Removable discontinuity occurs when both the numerator and denominator are zero

We will check this by inputting x = -4 in the expression, we have:

$\begin{gathered} f(x)=\frac{x^{2}-2x-24}{x+4} \\ x=-4 \\ f\mleft(x\mright)=\frac{-4^2-2\left(-4\right)-24}{-4+4} \\ f\lparen x)=\frac{16+8-24}{-4+4} \\ f\mleft(x\mright)=\frac{24-24}{-4+4} \\ f\mleft(x\mright)=\frac{0}{0} \end{gathered}$

Thus, when x = -4, we have a removable discontinuity

Hence, our answer is: (x, y) = (-4, 0)