Question

f(x)=(x+3)(6-x)/(x-2)(x+3). State the the x-coordinate of any holes in the graph of this function. I got 0. Is that right? Please check my work. Thank you.

Answer 1

Holes would occur whenever the denominator is zero... so when is the denominator
(x-2)(x+3) zero?

Answer 2

never?

Answer 3

(x-2)=0
(x-3)=0

Answer 4

Of course it can :)
For a product of two numbers
ab to be zero
ab = 0
You only need ONE of them (either a or b) to be zero.

Having said that
(x-2)(x+3) = 0
You only need one of these two factors to be zero :)

Answer 5

can I please get a visual?

Answer 6

Is it 2?

Answer 7

Okay, here's an example :)

(x - 20) (x + 13) = 0
You only need one of these two factors to be zero, so either
x - 20 = 0
OR
x + 13 = 0

So what are the two possibilities?

Answer 8

X=20 or x=-13

Answer 9

That's right :)
So, back to your denominator...
When, then, is
(x-2)(x+3) equal to zero?

Answer 10

x=2 or x=-3

Answer 11

And those... are when your function has holes ^.^

Answer 12

Okay. So would the answer be 2?

Answer 13

2, and -3
Remember, whenever x equals any of them (2 or -3) the denominator becomes zero, and the function cannot exist there, hence the hole.

Answer 14

okay. i understand. thank u. ^_^

Answer 15

No problem ^.^

Answer 16

WAIT
Are you still there?

Answer 17

yes.

Answer 18

I'm sorry, a hole is when BOTH the numerator and denominator are zero
So you also have to find when the numerator
(x+3)(6-x) =0
When is the numerator zero?

Answer 19

when they both are different?

Answer 20

No, do the same thing with the denominator, the way you found that it becomes zero when x=2 or x=-3

(x+3)(6-x) = 0

Answer 21

x=-3 or x=6

Answer 22

That's right
So the numerator is zero when x = -3 or x = 6
The denominator is zero when x = 2 or x = -3


When are they BOTH zero

Answer 23

i'm not sure.

Answer 24

@PeterPan i came up with -3

Answer 25

Yeah, that's correct :)

Answer 26

Thank you. ^_^

Answer 27

No problem :)