How do I go about solving this problem?

Question

For the function f(x) = x^2 / x^2 - 9

1. Vertical and horizontal asymptotes

2. Intervals increasing and decreasing

3. Intervals of concave up and down

4. Local maxima and minima, and points of inflection

5. Sketch a graph of f from the above information

Thanks

Answer 1

The function is constant, is it not? f(x) = 1 - 9 = - 8.[ isn't x^2/x^2 = 1?]

So nothing can be done.

Answer 2

1) for vertical asymptotes factor out denominator

x^2/(x-3)(x+3)

so vertical asymptotes are x= 3 and x=-3 since the graph has no finite values at these 2 points

for horizontal since the degree of both numerator and denominator are same. divide leading coefficient of numerator by leading coefficient of denominator so 1/1 is 1 horizontal asympototes is y= 1

2) so graph is obviously split into 3 parts one from x<-3 x-> - infinity one from x> -3 to x<3 and one from x>3  to x -> infinity

now to find intervals you just have to test out points using equations and see if they increase or decrease

for (-infinity, -3) the interval is increasing

for (-3,0] the interval is increasing

for [0,3) the interval is decreasing

for (3, infinity) the interval is decreasing

3) for concave up and down use second derivative test

calculate second derivative of function (extremely tedious)

f " (x) = 2/(x^2-9) - 10x^2/ (x^2-9)^2 + 8x^4/ (x^2-9)^3  

if f " (x) > 0 at point c then it is concave up at point c

if f " (x) < 0 at point c then it is concave down at point c

so test out points x < -3 we get negative so it is concave down

testing points from -3<x<or equal to 0 it is concave up

testing points from 0< x < 3 it is concave down

testing points x>3 it is concave up

4) graph has 3 parts local maxima between -3>x>3 is 0

and no parts have local minima

and now you can graph it with this info

hope this helps