Question

Anyone good with pre calc?

Answer 1

Can find restrictions on domain by setting denominator = 0. and solving for x. Those will be the values that would cause division by zero. Any other x in the domain would be allowed.

Answer 2

X- and Y-intercepts are standard procedure. Let y=f(x)=0 and solve for x to find x-intercepts, then let x=0 to solve for y-intercepts.

Answer 3

huh?
@CliffSedge

Answer 4

You're asking about this, right?

\(\large y=\frac{2}{x^2-2x-3}\)

Answer 5

yes

Answer 6

The best first step is to factor the bottom to find where x causes division by zero, that will inform your domain and tell you where the vertical asymptotes are.

Answer 7

so ( x + 1) ( x - 3) ?

Answer 8

Yes, when you set that =0. you'll see what x is not allowed to be (would cause division by zero, so is restricted from the domain).

Answer 9

Also, since there aren't like factors on top to cancel either one of those, the discontinuities will show up as vertical asymptotes and not just holes in the graph.

Answer 10

I forget how to work that out
would it be x= -1 and x=3 ?

Answer 11

Right, those are the equations of the vertical asymptote lines.
They are also what must be excluded from the domain.

Answer 12

so domain would be all real numbers except -1 and 3?

Answer 13

@CliffSedge

Answer 14

Exactly.

Answer 15

Alright. so to find the range?

Answer 16

I recommend finding the intercepts and horizontal asymptote first, then sketching the graph to get a better idea of what the range is.

Answer 17

Alright, how do i do that?

Answer 18

I'll repeat my second post, so you don't have to scroll up.

"X- and Y-intercepts are standard procedure. Let y=f(x)=0 and solve for x to find x-intercepts, then let x=0 to solve for y-intercepts."

Answer 19

so i do 0 = 2 / x^2 -2x - 3 ?

Answer 20

@CliffSedge

Answer 21

For x-intercepts, yes, but notice how it only matters if the numerator is equal to zero (which never happens...)

Answer 22

I dont understand how to do it though

Answer 23

To find the intercepts? The X-intercept would happen when f(x)=0, but since all you have is a 2 in the numerator, it doesn't matter what x is, y will never equal 0. There are no X-intercepts.

Answer 24

For the Y-intercept, do the reverse: plug in 0 for x and solve for y.

Answer 25

oh! thats what you were saying!

Answer 26

Yeah, sorry if the vocab is a little confusing. :">

Answer 27

no no its not your fault.
okay, so -2/3 is y? @CliffSedge

Answer 28

I don't know, I haven't worked it out yet . . . just a sec...

Answer 29

Yes, y=-2/3 is the y-intercept.

Answer 30

So, you have a Y-intercept but no X-intercept; that should give you a clue about the horizontal asymptote.

Answer 31

(You should already be sketching the graph by now)

Answer 32

lol, well ya see. I have no idea. How so i find that? if theres no x int that means there are no horizontal asymt. either? @CliffSedge

Answer 33

No, the horizontal asymptote is the long-term behavior, i.e. what happens as x approaches -∞ and ∞.

Answer 34

You should at least have an x-y table going by now, e.g.

Answer 35

still have no idea. @CliffSedge

Answer 36

Hmm, ok, well think about what would happen if x were some really really REALLY ridiculously large number, what would happen to y?

Answer 37

y would be stretched?

Answer 38

Im sooooo lost.. @CliffSedge

Answer 39

all i need is the range ans the horizontal asy.!

Answer 40

The numerator is just a number, the denominator contains an expression with x^2 in it. If x gets ridiculously large (in either the positive or negative direction), you'll eventually reach 2/∞ which goes to zero. It never actually touches zero or there would be an X-intercept there, instead, y=0 is a horizontal asymptote.

Answer 41

so then the range?

Answer 42

So we know that the curve never touches the x axis, but it does approach it asymptotically. You need to find out whether all the y-values are negative, positive, or both, if both, then there is no restriction on the range.

Answer 43

so all real numbers?

Answer 44

It looks like it, yes (almost). There is an interval where y is negative, but most of the time it is positive. Since there is no X-intercept, however, y cannot equal 0.

Answer 45

uhhh okay. so y doesnt = 0..?

Answer 46

????????? @CliffSedge

Answer 47

Right, when looking for the Y-intercept, you already found that y cannot equal zero because there is no value of x that will do that for you. The curve approaches y=0 asymptotically (the horizontal asymptote), but never actually touches it.

Answer 48

The graph looks kinda like this: