What are the coordinates of the hole in the graph of the function f(x)? f(x)=x^2+4x−12 / x−2

The coordinates are (2,8)

Step-by-step explanation:

A hole is where both the numerator and the denominator are zero

f(x)=x^2+4x−12 / x−2

Factor the numerator

f(x) = (x+6) (x-2)/ (x-2)

The hole will occur where x-2 =0

x-2=0

Add 2 to each side

x-2+2 =0+2

x=2

There is a hole at x=2

If we could cancel the x-2 values from the top and bottom, we are left with

f(x) = x+6

At x=2

f(2) = 6+2

f(2) would be 8

The coordinates are (2,8)

There is a hole

x = ± 6i

step-by-step explanation:

x^2 = -36

take the square root of each side

sqrt(x^2) = sqrt(-36)

we know sqrt(ab) = sqrt(a) sqrt(b)

x = ±sqrt(36) sqrt(-1)

we know that sqrt (-1) = i

x = ± 6i

fa tha first one you gonna add y to both sides

it look like this

x-y+y=6+y

x=y+6

therefore your answer is

x =y+6

fa tha second one

you gonna flip tha equation

it gonna look like x-4y=y

then add 4 to both sides

it gonna look like tha

x-4 +4 =y +4

so therefore tha answer is x=y+4

step-by-step explanation: