Calculus Help!!!?

Question

Label all x-intercepts, relative extrema, points of inflection, and asymptotes. I will give a few that I am having trouble with. I have beat my brains out trying to get them, for some reason, the teacher does not go over the difficult ones in class.

1.) f(x) = x + 32/x

2.) y=x sq.rt. of 16-x^2

3.) y= -1/3(x^3-3x+2)

Answer 1

1.) f(x) = x + 32/x
x=0 is a vertical asymptote
dy/dx = 1 -32/x^2
x^2 = 32
x = +/- 4sqrt(2) are the critical points
f(x) has a relative minimum at x = 4sqrt(2) and a relative maximum at x = -4sqrt(2).
Since f(x) = (x^2+32)/x the line y =2x is also an asymptote.There are no x- or y-intercepts.

2.) y=x sq.rt. of 16-x^2
If |x| > 4, f(x) is imaginary. Hence the graph of f(x) exists only in the range -4= The graph passes through the origin and thus has x- and y- intercepts of 0
dy/dx = 16-2x^2 so +/- x = 2sqrt(2) are critical points.
There is a local max at 2sqrt(2) and a local min at -2sqrt(2)

3.) y= -1/3(x^3-3x+2)
dy/dx - 1/3(3x^2 - 3) = -x^2+1 so there are critical points at x = +/- 1. There is a local minimum at x=-1 and local max at x=1.

The y- intercept is (0,-2/3) and x- intercept at1,0).
There are no asymptotes. (0,-2/3) is a point of inflection.

Answer 2

Use a graphing calculator, they'll find these for you no problem
-to find the x-intercepts, set y = 0, and solve for x
-to find relative extrema, take the derivative of the function, and plug in values between the given interval & endpoints
-to find asymptotes, set the numerator = 0 to find horizontal asymptotes, to find vertical asymptotes, set the denominator = to 0
-points of inflection, I'm not really sure