Question

describe the vertical asymptote(s) and hole(s) for the graph: y=(x-4)/(x^2+3x+2)

Answer 1

hint:
y=(x-4)/(x^2+3x+2) = (x-4)/ [(x+1)(x+2)]

Answer 2

I am honestly so lost, I have no idea what this means

Answer 3

Oh, do you have to factor to get that answer?

Answer 4

This is a situation in which your doing some online research is essential. You might want to Google "vertical asymptote" and learn that way what such an asymptote is and why we bother to find it (or them).

Look at just what sourwing factored: the denominator (only). We are to find the values of x that make the denominator (only) = zero (0). Factoring the den. makes this easier.

Take sourwing's two factors and set them equal to zero separately, and then solve for x in each case. Please type out your answers in the form x=a, x=b.

Answer 5

x=-1 and x=-2

Answer 6

But how do I figure out if, and where there are holes

Answer 7

First: by typing out x=-1 and x=-2, you have correctly identified the two vertical asymptotes.

Regarding "holes:" Look at the numerator of the given rational function you started with. What is the numerator?

Answer 8

x-4

Answer 9

Is x-4 equal to either of the two factors of the denominator?

Answer 10

No, so is the hole -4, or 4?

Answer 11

I don't really understand what a "hole" is, mathematically speaking

Answer 12

"equal" = "the same as"

Right. No. Therefore you will not be canceling any factors, right? You can't cancel x-4 and x=-1.

Answer 13

Now suppose that your numerator were x-4 and the denom. were (x-4)(x+2). Could you cancel anything?

Answer 14

the fours?

Answer 15

because you would come up with a positive four in the denominator

Answer 16

Not exactly, but you could cancel the FACTORS (x-4) and (x-4) in numerator and denom. respectively.

Answer 17

No, you cancel only if the factors are exactly the same; You can cancel (x-4) / (x-4), but not (x-4) / (x+4).

Answer 18

Oh, ok.

Answer 19

So, in my example, can you, or can you not, cancel?

Answer 20

you can't

Answer 21

Let me write out my example again, for clarity:\[y=\frac{ x-4 }{ (x-4)(x+2) }\]

Answer 22

Can you do any cancelling here?

Answer 23

yes the x-4 and x-4 so you'd end up with just y=x+2

Answer 24

In other words, y ou'd end up with \[y=\frac{ 1 }{ 1(x+2) }.\]

Answer 25

Now, because the original function had (x-4) in both numerator and denominator, x could NOT be 4. Why not?

Answer 26

Because you cancelled out the equation where x would equal 4, so it's not possible.

Answer 27

Not quite. The reason is that if you have (x-4) in the denom., and then let x=4, you'll have zero in the denom. Right?

Answer 28

yes

Answer 29

Not allowed. THAT is where your graph would have a hole!!

You'd have a nice, smooth graph for x>-1, BUT still, you could not let x=4, because then the factor (x-4) in the original problem statement would be zero. Division by zero is not permitted. So, instead of drawing a nice, smooth graph right past x=4, you'd have to draw a hole in the graph where x=4.

Answer 30

Oh! Okay, that makes sense

Answer 31

So, in summary, your graph will have a hole in it if you have a factor in the denom. equal to a factor in the numerator. Even tho' you can cancel, you are not changing the fact that x can not equal 4 in the original function.

Answer 32

u r awesome. Happy for the chance to work with you!

Answer 33

Thank you!

Answer 34

My pleasure! Hasta la vista!