that's right. so the vertical asymptotes are x = -1 and x = -2

for the horizontal asymptotes, find and . if you get a finite answer for either (in this case, they will both give the same answer), that's your horizontal asymptotes. it is a line y = something

don't know what you mean. this is finding the x- (and y-) interceptAxis cross should be x = 0 and y = 0. It is (0,0) so it should be solved like that.

for turning points, you only need the first derivative, and set it to zero.Curve's turning points. I want to see if differentiating should be enough, or do i need to go down to and make it =0. If the answer is imaginary it's proved, correct?

yup, is your graph correct?sketching curve should be done using the information gained from the previous questions

huh? oh, ok, i think i got what you're saying. um, yeah, estimates are good when graphing.and finding out which side the curve goes when approaching the asymptotes (asy+0.0001), (asy - 0.00001) to get a quick estimate.

nope, you reflect in the y-axisf(-x) means a reflection in the x axis? Thanks for the posts