Question

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)

Answer 1

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.

Answer 2

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

Answer 3

@zepdrix please help

Answer 4

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

Answer 5

yeah i used the quotient rule

Answer 6

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

Answer 7

yeah the first answer was correct.

Answer 8

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 . :(

Answer 9

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.

Answer 10

right but wouldn't it be false?

Answer 11

1=0? doesn't work.

Answer 12

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?

Answer 13

by undefined what do you mean?

Answer 14

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

Answer 15

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

Answer 16

like this one?

Answer 17

oh ya 2!

Answer 18

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?

Answer 19

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

Answer 20

no.

Answer 21

x cannot equal to 2.

Answer 22

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

Answer 23

all real numbers except 2.

Answer 24

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

Answer 25

so theres no relative extrema?

Answer 26

yes last but not least, inflection points. :)

Answer 27

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

Answer 28

no but let me try real quick

Answer 29

so inflection points is always SECOND DERIVATIVE right?

Answer 30

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.

Answer 31

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

Answer 32

@zepdrix

Answer 33

Dude dude dude -_-
what..?

Don't expand out squares, especially if they're in the denominator.

Answer 34

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

Answer 35

yeah i did:/

Answer 36

i used the rule book.

Answer 37

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

Answer 38

ohhhhh yes

Answer 39

crap.

Answer 40

XD

Answer 41

-2/(x-2)^3

Answer 42

ok :D

Answer 43

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

Answer 44

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

Answer 45

huh?

Answer 46

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

Answer 47

so threres no inflection point?

Answer 48

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.

Answer 49

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

Answer 50

no it didn't...

Answer 51

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

Answer 52

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

Answer 53

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

Answer 54

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.. :\

Answer 55

Oh the other curve, umm

Answer 56

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