Thread: Have to Graph This!

Ok I'm supposed to find the asymptotes, the critical numbers, the inflection points for this function. I tried to solve for the x-int and everything went downhill from there.

y=2(cosx)^2 - sinx

I got my x-int to be (1+-sqrt(17))/-4 lol I don't know if thats right at all.

I put 2(1-(sinx)^2) - sinx = 0 then tried to use the quadratic formula to solve for x. Came out ugly

Got my y-int as 2, just set the x = 0 ...

y' was 4(cosx)(-sinx) -cosx but i couldnt solve that = 0 to find the critical numbers... I think two of them are pi/2 and -pi/2 but... yeah. I'm stuck.

Any help would be GREATLY appreciated. Thank you so much.

Originally Posted by drain

Ok I'm supposed to find the asymptotes, the critical numbers, the inflection points for this function. I tried to solve for the x-int and everything went downhill from there.

y=2(cosx)^2 - sinx

I got my x-int to be (1+-sqrt(17))/-4 lol I don't know if thats right at all.

I put 2(1-(sinx)^2) - sinx = 0 then tried to use the quadratic formula to solve for x. Came out ugly

Got my y-int as 2, just set the x = 0 ...

y' was 4(cosx)(-sinx) -cosx but i couldnt solve that = 0 to find the critical numbers... I think two of them are pi/2 and -pi/2 but... yeah. I'm stuck.

Any help would be GREATLY appreciated. Thank you so much.

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

I can't believe he made this so messy... thanks Jhevon

Originally Posted by drain

I can't believe he made this so messy... thanks Jhevon

yes, it is complicated, but thank goodness you don't have to graph it. once you get the complicated numbers, you're done

just in case you're curious, here's what the graph looks like. good luck graphing that thing!