beobachtereb

2022-09-25

How do you find the zeros for the function $f\left(x\right)=\frac{{x}^{2}-x-12}{x-2}$?

AKPerqk

Beginner2022-09-26Added 9 answers

$f\left(x\right)=\frac{{x}^{2}-x-12}{x-2}$

$=\frac{(x+3)(x-4)}{x-2}$

$\therefore f\left(x\right)=0\to \frac{(x+3)(x-4)}{x-2}=0$

Hence the zeros of f(x) occur when (x+3)(x−4)=0

Note: f(x) is undefined at x=2

$\therefore f\left(x\right)=0$ at x=−3 and x=4

This can be seen from the grapg of f(x) below:

graph{((x+3)(x-4))/(x-2) [-18.01, 18.04, -9.01, 8.99]}

$=\frac{(x+3)(x-4)}{x-2}$

$\therefore f\left(x\right)=0\to \frac{(x+3)(x-4)}{x-2}=0$

Hence the zeros of f(x) occur when (x+3)(x−4)=0

Note: f(x) is undefined at x=2

$\therefore f\left(x\right)=0$ at x=−3 and x=4

This can be seen from the grapg of f(x) below:

graph{((x+3)(x-4))/(x-2) [-18.01, 18.04, -9.01, 8.99]}

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