Answer:
Tangent at (1,2) has value 
. And (-2, 4) is the point of intersection of horizontal tangent.
Step-by-step explanation:
Given curve equation,
To find tangent at (x,y)=(1,2) and point of intersection of horizontal tangent, differentiate (1) withrespect to x we get,
At (1, 2), tangent line is,
To find point of intersection of horizontal tangent we have to do,
Thus,
At x=0, y=0
but snce,
at (0,0) there exist a vertical tangent. And,
At x=-2, y=4.
Thus (-2, 4) is the point of intersection of horizontal tangent.
Answer:
ιт ιѕ (в.) -2χ(χ+3)(χ+2)
Step-by-step explanation:
нσρє ι ¢συℓ∂ нєℓρ уσυ συт тσ∂αу!!!!!!!!
Answer: sin u = -5/13 and cos v = -15/17
Step-by-step explanation:
The nice thing about trig, a little information goes a long way. That’s because there is a lot of geometry and structure in the subject. If I have sin u = opp/hyp, then I know opp is the opposite side from u, and the hypotenuse is hyp, and the adjacent side must fit the Pythagorean equation opp^2 + adj^2 = hyp^2.
So for u: (-5)^2 + adj^2 = 13^2, so with what you gave us (Quad 3),
==> adj of u = -12 therefore cos u = -12/13
Same argument for v: adj = -15,
opp^2 + (-15)^2 = 17^2 ==> opp = -8 therefore sin v = -8/17
The cosine rule for cos (u + v) = (cos u)(cos v) - (sin u)(sin v) and now we substitute: cos (u + v) = (-12/13)(-15/17) - (-5/13)(-8/17)
I am too lazy to do the remaining arithmetic, but I think we have created a way to approach all of the similar problems.
