a function is increasing on the interval where its derivative is positive, and decreasing where it is negative

these are the points where either the derivative is zero, or the function is undefinedcritical points

these occur where the derivative is zero. they can be maximums, minimums or inflection points. the second derivative test can tell you which.relative extrema

a function is concave up on the interval its second derivative is positive-concavity and points of inflection

a function is concave down on the interval its second derivative is negative

if the second derivative is zero at a point, then it MIGHT be an inflection point. further tests are needed.

you can test for an inflection point in one of to ways:

(1) using thefirstderivative: the the sign of the derivative changes on either side of the said point, thenit is notan inflection point. test points very close to the point you are considering.

(2) using thesecondderivative: if the second derivative changes sign on either side of the point, then itisan inflection point

vertical asymptotes occur where the function is undefined.-Vertical, Horizontol, and slant asymptotes

horizontal asymptotes occur if either of the following limits are finite. if they are finite, then the horizontal asymptote is the value of the limit

and

you have to do both limits. you can have up to two horizontal asymptotes

obviously we are not looking at the same function here, or you typed something wrong. the function above is a polynomial, there are no asymptotes.-use the info to sketch the curve, use tables for f' and f''

1.x^4 - 4x^3 +16x

I found the vertical asymptote is 2.. But how do I find the slant asymptote? There's no other factor to use long division to find it..

right for the HA. for the VA, you are wrong. there are two points where2.3x^(2/3) - 2x

4. (x^2+1)/(x^2-2)

I found the H.A. is 1.. How do I find the V.A(The denominator is never zero..)