Question

refresher on how to calculate vertical and horizontal asymptote

x^2/x - 1

Answer 1

Vertical asymptote is 1 because 1-1=0 and the denominator can never be zero.

The horizontal asymptote is found as x=>inf, so it would be Inf which means it is straight up on the graph.

Answer 2

Wait, not straight up, since it is horizontal.

Answer 3

what are the critical points

Answer 4

Derivative set equal to zero

Answer 5

if the derivative of this function is

(x-2) x/ (x-2)^2

then the critical points are?

Answer 6

Different equation?

Answer 7

the derivative of x^2/x-1 is what i just posted, no?

Answer 8

Set it equal to zero. You get x=0,2

Answer 9

oh ok

Answer 10

where is it increasing and decreasing...local maximum and minimum?

Answer 11

That would be the critical points. Plug them into the second derivative and if you get a negative answer, then it is a max and if it is positive, then it is a min.

Answer 12

whoa...what if i get all positives, how will i determine local minimum?

Answer 13

You will get a negative, don't worry. :/

Answer 14

lol

Answer 15

that was for max and min right?

what about increasing and decreasing

Answer 16

Inflection points are found when the second derivative is set equal to zero.

Answer 17

and concaves, up or down?

Answer 18

If the second derivative with the critical points plugged in is negative, then it is concave down and positive is concave up

Answer 19

where the function is increasing or decreasing

Answer 20

This graph actually has an oblique asymptote, so I'm not sure how your graph looks, but it when he increasing as it goes up against the vertical asymptote.

Answer 21

thanks