Question

I need quick help!

Answer 1

I feel like there is a missing graph or something?

Answer 2

@Hero

Answer 3

@whpalmer4

Answer 4

well, yes, we certainly don't have all the information here...is this part of another problem?

Answer 5

That's everything it gives me.... That's why I'm so confused. I think the equation is the formula itself..

Answer 6

there's no unique solution to that problem, unless there are additional constraints provided. You can make an infinite number of equations that are a constant divided by the variable \(w\). Unless it has to go through some point, or match some graph, or be equal to some other expression, there's no way of knowing what the value of the constant should be.

Answer 7

Without that, it's the equivalent of asking you to guess the number I'm thinking of, with about the same likelihood of success :-)

Answer 8

Yeah that's what I was thinking.. /: So what answer should I put down? Haha. Nothing?

Answer 9

I'd have another look and make sure that it isn't somehow related to the previous problem on the page...

Answer 10

Okay.. /:

Answer 11

Am I correct?

Answer 12

@whpalmer4

Answer 13

let's find out:

\[\frac{1}{x} -\frac{1}{x+1} = \frac{1}{x(x+1)}\]\[\frac{1}{x}*\frac{(x+1)}{(x+1)} - \frac{1}{(x+1)}*\frac{x}{x} = \frac{1}{x(x+1)}\]\[\frac{x+1-x}{x(x+1)} = \frac{1}{x(x+1)}\]\[\frac{1}{x(x+1)} = \frac{1}{x(x+1)}\checkmark\]
looks good!

Answer 14

Am I right?

Answer 15

we could also test it out by substituting in some numbers for \(x\) and making sure it works:

\[\frac{1}{2} - \frac{1}{2+1} = \frac{1}{2(2+1)}\]\[\frac{1}{2} - \frac{1}{3} = \frac{1}{6}\]\[\frac{3}{6} - \frac{2}{6} = \frac{1}{6}\checkmark\]
\[\frac{1}{3}-\frac{1}{3+1} = \frac{1}{3(3+1)}\]\[\frac{4}{12} - \frac{3}{12} = \frac{1}{12}\checkmark\]

Answer 16

Not correct for the next one. To have vertical asymptotes, you must have an equal number of unique values of the independent variable where the denominator becomes 0.

Answer 17

So c?

Answer 18

as another check, try evaluating the function at x = 10 and see if the result you get looks like the value shown on the graph

Answer 19

I'm not sure how to do that..

Answer 20

No, c is not correct either — it only has one value of x where the denominator equals 0. Your graph requires the denominator to equal 0 in two places, at x = 0 and at x = 3...do you see that?

Answer 21

Yes I see that.

Answer 22

Right?

Answer 23

looks reasonable