anonymous
  • anonymous
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)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • 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
  • anonymous
i tried to also find the derivative and i got 1/ (x-2)^2
anonymous
  • anonymous
@zepdrix please help

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

zepdrix
  • zepdrix
So did you apply the Quotient Rule? One sec, I'll check it :O
anonymous
  • anonymous
yeah i used the quotient rule
zepdrix
  • zepdrix
Mmm yah I think your first answer was correct: 1/ (x-2)^2
anonymous
  • anonymous
yeah the first answer was correct.
anonymous
  • 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
  • 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
  • anonymous
right but wouldn't it be false?
anonymous
  • anonymous
1=0? doesn't work.
zepdrix
  • 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
  • anonymous
by undefined what do you mean?
zepdrix
  • zepdrix
Numbers that would cause a problem :D Like dividing by zero.
zepdrix
  • zepdrix
Any x values that would cause a problem for our derivative?
anonymous
  • anonymous
like this one?
anonymous
  • anonymous
oh ya 2!
zepdrix
  • 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
  • zepdrix
You stated the domain earlier, so go back and look at it! :)
anonymous
  • anonymous
no.
anonymous
  • anonymous
x cannot equal to 2.
zepdrix
  • zepdrix
Good good. Since it's not in the domain of our function, it can't be a critical point.
anonymous
  • anonymous
all real numbers except 2.
zepdrix
  • zepdrix
So there are no critical points. Should we look for inflection points next? :o
anonymous
  • anonymous
so theres no relative extrema?
anonymous
  • anonymous
yes last but not least, inflection points. :)
zepdrix
  • zepdrix
So we'll need to find the second derivative for that. Have you tried that step yet?
anonymous
  • anonymous
no but let me try real quick
anonymous
  • anonymous
so inflection points is always SECOND DERIVATIVE right?
zepdrix
  • 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
  • anonymous
i got (-3x^3+17x^2-32x+20)/(x^2-4x+4)
anonymous
  • anonymous
@zepdrix
zepdrix
  • zepdrix
Dude dude dude -_- what..? Don't expand out squares, especially if they're in the denominator.
zepdrix
  • zepdrix
Did you apply the quotient rule? We don't want to do that.
anonymous
  • anonymous
yeah i did:/
anonymous
  • anonymous
i used the rule book.
zepdrix
  • zepdrix
\[\Large y'\quad=\quad \frac{1}{(x-2)^2}\quad=\quad (x-2)^{-2}\]Power rule from there :O
anonymous
  • anonymous
ohhhhh yes
anonymous
  • anonymous
crap.
zepdrix
  • zepdrix
XD
anonymous
  • anonymous
-2/(x-2)^3
anonymous
  • anonymous
ok :D
zepdrix
  • zepdrix
Ok cool. Setting our second derivative equal to zero:\[\Large 0\quad=\quad \frac{-2}{(x-2)^3}\]
zepdrix
  • zepdrix
Hmm I think inflection points work the same way as critical points ~ they exist where the second derivative is undefined.. mmmmm
anonymous
  • anonymous
huh?
anonymous
  • anonymous
don't we plug in severals x's between the intervals or something?
anonymous
  • anonymous
so threres no inflection point?
zepdrix
  • 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
  • zepdrix
Oh the problem didn't ask for concavity did it? -_- oh.. hmm
anonymous
  • anonymous
no it didn't...
anonymous
  • anonymous
so help me with inflection point, do we have to include other x's between the intervals?im not sure
zepdrix
  • zepdrix
include other x's... whu? +_+
anonymous
  • 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
1 Attachment
zepdrix
  • 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
  • zepdrix
Oh the other curve, umm
zepdrix
  • zepdrix
bah i dunno >:O my head hurts.. i need a maf break!

Looking for something else?

Not the answer you are looking for? Search for more explanations.