Known curves C y x 3 3x 2 2x

1.Tangent slope k=y'=3*x 2-6*x+2 when k'=0, k is the minimum value.

k'=6*x-6=0

x=1k=-1

Tangent (1,0).

The tangent equation y=-1*(x-1)+0=1-x

2.The slope of the tangent line passing through the origin k=y x=x 2-3*x+2 corresponding to the tangent coordinates should meet k=y'

x 2-3*x+2=3*x 2-6*x+2x=3 2 tangent coordinates (3 2, -3 8).

Tangent equation y=3 2*x

1.I didn't learn the countdown.

k=y`=3x^2-6x+2

3(x-1)^2-1

k is minimum -1, where x = 1 and y = 0

The tangent equation is y=-x+1

2.Because (0,0) is on the curve c.

So k = 3 (0-1) 2-1 = 2

y=2x

1) Slope y'The minimum value of =3*x 2-6*x+2 is let y''=6*x-6=0, we get x=1, so the slope y'=-1, y=0, and the tangent equation is y=-(x-1).

2) Lingy'=3*x 2-6*x+2=y x=x 2-3x+2, 2*x 2-3x=0, x=, tangent coordinates can be obtained, and then the slope can be obtained to obtain the tangent equation.

The integral of the curve y=(2 3)x (3 2) integral can be used by the formula l=[(1+y'2) (1 2)].

Because. l= rotten branch [(1+y.)]'^2)^(1/2)]dxy'=x^(1/2)

So y=(2 3)x (3 2).

l= Hungry Spikemin[(1+x) (1 2)]dxl=2 Clan 3*(1+x) (3 2).

1) Slope y'The minimum value of =3*x 2-6*x+2 is let y''=6*x-6=0, we get x=1, so the slope y'=-1, y=0, and the tangent equation is y=-(x-1).

2) Lingy'=3*x 2-6*x+2=y x=x 2-3x+2, 2*x 2-3x=0, x=, tangent coordinates can be obtained, and then the slope can be obtained to obtain the tangent equation.

1.Tangent slope k=y'=3*x 2-6*x+2 when k'=0, k is the minimum value.

k'=6*x-6=0

x=1k=-1

Tangent (1,0).

The tangent equation y=-1*(x-1)+0=1-x

2.The slope of the tangent line passing through the origin k=y x=x 2-3*x+2 corresponding to the tangent coordinates should meet k=y'

x^2-3*x+2=3*x^2-6*x+2x=3/2

Tangent coordinates (3 2, -3 8).

Tangent equation y=3 2*x

Hello! 1.I didn't learn the countdown.

k=y`=3x^2-6x+2

3(x-1)^2-1

k is minimum -1, where x = 1 and y = 0

The tangent equation is y=-x+1

2.Because (0,0) is on the curve c.

So k = 3 (0-1) 2-1 = 2

y=2xIf in doubt, please ask.

The line l is tangent to the curve c at the point (x0, y0) = the function value is equal: x0 3 - 3x0 2 +2x0 =kx0=> x0 2 - 3x0 + 2 =k and the slope is equal 3x0 2 - 6x0 +2 =k =>x0=3 2 =>k=-1 4 y0=-3 8

The derivative of the original function y=3x 2+3 makes y=15 and x= 2 in a two-point formula.

y=15x+16 or y=15x-16

y'=3x 2-1,x=1 generation fiber ease-in,get,y'=2, that is, the slope of the tangent line at the point is 2, let the equation be y=2x+c, and substitute the coordinates of p(1,2), that is, guess the vertical elimination: 2=2+c, and obtain: c=0, so the tangent equation is:

with y=2x

The derivative of the original function y=3x 2+3 makes y=15 and x= 2 in a two-point formula.

y=15x+16 or y=15x-16

y'=3ax^2+2bx+c

y"=6ax+2b

Points (1, -10) are inflection points.

So 0=6a+2b

x=-2 is the coarse buried site.

So 12a-4b+c=0

(1,-10) and (-2,44)10=a+b+c+d

44=-8a+4b-2c+d

a=1,b=-3,c=-24,d=16

p should be brought into the straight line, n=3 should be brought in, and then the hyperbola should be brought in to find m=-2

For ab two ants to lift only the point of the words stuffy, hyperbolic scramble line in x