X-int solve for

Y-int

For the others take the derivative, set up test intervals based upon the zeros of the derivative, places it is undefined, and places the original function is undefined, test an element of each interval in the first derivative, a change from positive to negative denotes a max and from negative to positive a min, and then obviously the intervals that are positive are decreasing and those that are negative are decreasing. Graph it, and for inflection points find the second derivative, find its zeros, and points of discontinuity, now apply the same concept as finding max and mins except whenver there is a sign change it is an inflection point, and negative intervals are decreasing, positive increasing