PLEASE HELP :( Please help: sketch the graph of a function. label the intercepts, relative extrema, points of inflection, and asymptotes! then state the domain of the function! y= (5-3x)/(x-2)

- anonymous

- jamiebookeater

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- anonymous

so i found the vertical asymptote: x=2 and horizontal asymptote: y=3 and the intercepts i found that it is (0, -2.5) and (1.67,0)
so the domain is: (-inf,2)(2, inf)
now I'm stuck on how to do the relative extrema and the points of inflections. :( please help.

- anonymous

i tried to also find the derivative and i got 1/ (x-2)^2

- anonymous

@zepdrix please help

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- zepdrix

So did you apply the Quotient Rule?
One sec, I'll check it :O

- anonymous

yeah i used the quotient rule

- zepdrix

Mmm yah I think your first answer was correct: 1/ (x-2)^2

- anonymous

yeah the first answer was correct.

- anonymous

i mean I'm trying to find the relative extrema but i forgot how to. i know i have to find the derivative first. then I'm not sure where to go . :(

- zepdrix

So we have our first derivative:\[\Large y'\quad=\quad \frac{1}{(x-2)^2}\]
To find extrema/critical points, we set the first derivative equal to zero and solve for x.

- anonymous

right but wouldn't it be false?

- anonymous

1=0? doesn't work.

- zepdrix

Oh there is another little tid bit that I often forget.
critical points also exist when the first derivative is `undefined`.
Do you see any points where that might be true?

- anonymous

by undefined what do you mean?

- zepdrix

Numbers that would cause a problem :D Like dividing by zero.

- zepdrix

Any x values that would cause a problem for our derivative?

- anonymous

like this one?

- anonymous

oh ya 2!

- zepdrix

Yes! Good. x=2 is something we might consider to be a critical point.
Before we can say that for certain though, is x=2 in the domain of our function?

- zepdrix

You stated the domain earlier, so go back and look at it! :)

- anonymous

no.

- anonymous

x cannot equal to 2.

- zepdrix

Good good.
Since it's not in the domain of our function, it can't be a critical point.

- anonymous

all real numbers except 2.

- zepdrix

So there are no critical points.
Should we look for inflection points next? :o

- anonymous

so theres no relative extrema?

- anonymous

yes last but not least, inflection points. :)

- zepdrix

So we'll need to find the second derivative for that.
Have you tried that step yet?

- anonymous

no but let me try real quick

- anonymous

so inflection points is always SECOND DERIVATIVE right?

- zepdrix

y=0 gives us roots of the function.
y'=0 gives us critical points of the function.
y''=0 gives us inflection points of the function.
yes.

- anonymous

i got (-3x^3+17x^2-32x+20)/(x^2-4x+4)

- anonymous

@zepdrix

- zepdrix

Dude dude dude -_-
what..?
Don't expand out squares, especially if they're in the denominator.

- zepdrix

Did you apply the quotient rule?
We don't want to do that.

- anonymous

yeah i did:/

- anonymous

i used the rule book.

- zepdrix

\[\Large y'\quad=\quad \frac{1}{(x-2)^2}\quad=\quad (x-2)^{-2}\]Power rule from there :O

- anonymous

ohhhhh yes

- anonymous

crap.

- zepdrix

XD

- anonymous

-2/(x-2)^3

- anonymous

ok :D

- zepdrix

Ok cool.
Setting our second derivative equal to zero:\[\Large 0\quad=\quad \frac{-2}{(x-2)^3}\]

- zepdrix

Hmm I think inflection points work the same way as critical points ~ they exist where the second derivative is undefined.. mmmmm

- anonymous

huh?

- anonymous

don't we plug in severals x's between the intervals or something?

- anonymous

so threres no inflection point?

- zepdrix

between what intervals though?
We would need to find inflection points to determine which intervals to check.
I guess we just check where the function has a discontinuity since we have no inflection points.
So ya, check on the left side of 2 and also on the right.
We should get different concavity on each side for this problem.

- zepdrix

Oh the problem didn't ask for concavity did it? -_- oh.. hmm

- anonymous

no it didn't...

- anonymous

so help me with inflection point, do we have to include other x's between the intervals?im not sure

- zepdrix

include other x's... whu? +_+

- anonymous

so this is the answer: it doesn't say which is for which but i got the idea of what is what. my question is how'd they get the other curve on the right side

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- zepdrix

So they labeled the intercepts (as you found),
the asymptotes x=2, y=-3,
and nothing else...
So I think we did everything correctly, seeing as how we found no extrema for this one and they didn't label any on their graph.. :\

- zepdrix

Oh the other curve, umm

- zepdrix

bah i dunno >:O my head hurts.. i need a maf break!

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