i'm about to log off and don't have the time to do the entire problem.

the graph has no asymptotes, since the domain is not limited in anyway, the function is defined everywhere.

we find the critical points by finding y' and setting it to zero. we find the infleciton points by finding the second derivative and setting it to zero (or plug in the critical numbers you got before into the second derivative and seeing which one gives zero).

y = 2(cosx)^2 - sinx

for the y-int, x = 0

=> y = 2(cos(0))^2 - sin(0) = 2

for the x-int, y = 0

=> 0 = 2(cosx)^2 - sinx

=> 0 = 2(1 - (sinx)^2) - sinx

=> 0 = -2(sinx)^2 - sinx + 2

=> 2(sinx)^2 + sinx - 2 = 0

by the quadratic formula:

sinx = [-1 +/- sqrt(1 + 16)]/4

=> sinx = [-1 +/- sqrt(17)]/4

=> x = sin^-1{[-1 +/- sqrt(17)]/4}

yeah, that is messy. use your calculator for this, and remember this function will be periodic so you should add a + 2kpi or +kpi when done, depending on how it works out

y = 2(cosx)^2 - sinx

=> y' = -4cosxsinx - cosx

for critical points, set y' = 0

=> -4cosxsinx - cosx = 0

=> cosx(-4sinx - 1) = 0

=> cosx = 0 or -4sinx - 1 = 0

=> x = pi/2 + kpi for k some integer

or sinx = -1/4

and you can find the solutions for that

y'' = 4(sinx)^2 - 4(cosx)^2 + sinx

=> y'' = -4cos(2x) + sinx

so that should get you started, i'm outtie, good luck