Question

Find the domain, intercepts, the intervals where function is increasing and decreasing, extreme values, intervals of concavity, points of inflection, asymptotes (vertical, horizontal, or slant). Show the1st and the 2nd derivative tables. Sketch a graph.
(x^2 +12)/(x-1)

Answer 1

hmmm... looks familiar...

Answer 2

haha yeah after we stopped i was totally lost

Answer 3

and now i am like totally stressing because all this is due in 2 hours and i dont know if i can finish

Answer 4

\[\huge f(x)=\frac{x^2+12}{x-1} \]
\[\huge f'(x)=\frac{x^2-2x-12}{(x-1)^2} \]
i guess that's where we left off...

Answer 5

yupp

Answer 6

let's do all the things that don't require calculus first...domain, intercepts, asymptotes...

Answer 7

ok

Answer 8

look at f(x)... is there any x where f(x) is NOT defined?

Answer 9

x=1

Answer 10

ok... good... do domain = {all real except x=1}
this by the way is also where your vertical asymptote occurs... at x=1 is your vertical asymptote.

Answer 11

in a rational function, whatever makes the denominator = 0 is a vertical asymptote.

Answer 12

ok

Answer 13

now let's look for x and y intercepts...

Answer 14

lol why new question?

Answer 15

because I need as much help as a can get! ha

Answer 16

but it was sill open? I wrote a bunch of stuff for you.

Answer 17

to find the x-intercept, set y=0 so you need to solve:
\[\huge 0=\frac{x^2+12}{x-1} \]

Answer 18

what it was? im sorry i will try to go find it and look at that along with this, i dont know how to use this website very well

Answer 19

do you have a solution to this equation?

Answer 20

square root of -1/12?

Answer 21

i think you mean sqrt(-12) ?

Answer 22

if you say so :) i was kind of confused on how to solve that one

Answer 23

the only thing that matter is that numerator... if that numerator is 0, then it doesn't matter what the denominator is because you'll have 0 over some number... which is zero...

Answer 24

so you're actually solving this: \(\large 0=x^2+12 \)

Answer 25

oh ok

Answer 26

now, is the solution to that equation a real number?

Answer 27

no

Answer 28

so what does that tell you about the x-intercept(s)?

Answer 29

there arent any?

Answer 30

correct...

Answer 31

now let's look for the y-intercept..

Answer 32

what do i do to find that?

Answer 33

just set x=0 and figure it out f(0)

Answer 34

using the equation from earlier?

Answer 35

the function...

Answer 36

oh oh ok

Answer 37

-12?

Answer 38

so your y-intercept is at y=-12...

Answer 39

yes..

Answer 40

ok.. 1 more precalculus thing before we do the calculus...

Answer 41

\[\huge f(x)=\frac{x^2+12}{x-1} \]
in precalculus if the degree of the numerator = the degree of the denominator that you have a horizontal asymptote... do you have one here?

Answer 42

the numerator is greater than the demonimator

Answer 43

so do we have a horizontal asymptote?

Answer 44

no

Answer 45

if the degree of the numerator is 1 more than the degree of the denominator, then you have a slant asymptote.... as you can see, we have a slant asymptote... agreed?

Answer 46

yupp

Answer 47

do you know how to get the equation of the slat asymptote?

Answer 48

nope

Answer 49

have you ever worked with slant asymptotes?

Answer 50

not yet

Answer 51

that i can remeber anyways

Answer 52

to get the equation of the slant asymptote, all you have to do is do LONG DIVISION..... YAY!!!

Answer 53

oh ok