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

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

Step-by-step explanation:

We were given the function:

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

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

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