Using Graphing stategy find: Y=X/X^2-9?

Question

Find: Domain.Range.X int,Y int. Asymtotes, local extrema.Concavity. and if you can try to graph it

Answer 1

I just taught this section today to my calc students. You start by finding the first and second derivative. I recommend looking in your book for an example. Your book probably has something like this towards the front of the section. It would be too long winded to write everything down here. If you don't have your book on you, the web page below has problems that you can do and then check your answer.

http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/graphingdirectory/Graphing.html

Answer 2

In addition to ‘beanie_boy’!
lim(y)[x->-3]=-infinity, thus x=-3 is vertical asymptote on both sides;
lim(y)[x->+3]=+infinity, thus x=+3 is vertical asymptote on both sides;
while lim(y)[|x|->infinity]=0, thus y=0 is horizontal asymptote above and below of X-axis.
y(x) is odd function, as y(x)=-y(-x), thus symmetrical with point [0,0]
y’(x)=[1*(x^2-9)-x*2*x]/(x^2-9)^2=-(x^2+9)/(x^2-9)^2 is never =0, thus no extremes;
y’’(x)=-[2*x(x^2-9)-(x^2+9)*2*x]/(x^2-9)^4=18*x/(x^2-9)^4 is =0 if x=0.
y’’<0 for x=(-inf until –3) convex; y’’<0 for x=(-3 until 0) also convex;
y’’>0 for x=(0 until +3) concave; y’’>0 for x=(+3 until +inf) also concave;
point [0,0] being the cusp.
Thus it looks like hyperbola (1/x) with like tg(x) in between “hyperbola’s” branches!

Answer 3

is the question actually

Y = X/ (X^2 -9)?

asuming so

domain = all values such that denomination isnt = 0
so where X is not equal to 3 or -3

Range= -infinity to +infinity

X-intercept at X=0
Y-intercept at Y=0

U = x^2 -9
dU= 2XdX
Y' = 1/(2U) dU

Y'' = 1/2ln U dU

I believe, my calc is abit rusty

hope its a good start