Derivation is the slope of a curve at a certain point.

y'=3x？ + 1

Parallel to the straight line y=4x- 1, with a slope of 4.

So y'=3x? + 1=4，x 1 = 1； x2=- 1

Substitute the curve f(x)=x? +x-2, x 1 hour y 1=0.

Get Y 1=-4 when x2=- 1.

The coordinates of point p0 are (1, 0) or (-1, 0).

2。 Line 2x-6y+ 1=0 Slope = 1/3.

The slope of the straight line is -3.

And the curve y=x? +3x？ -1 tangency

Derivation is the slope of a curve at a certain point.

y'=3x？ +6 times

When y'=3x? +6x=-3 is x? +2x+ 1=0

Solve x=- 1 and substitute x=- 1 into y=x? +3x？ -1 gives y= 1.

So the coordinates of the tangent point are (-1, 1).

So the equation of tangent is y=-3x-2.

3 curve C:y=x? -2ax？ +2ax

Derive y'=3x? -4a+2a

Because the inclination of the tangent at any point is acute, that is, the slope is >: 0.

So: y'=3x? -4a+2a >0

3x？ -4a+2a=3(x？ -4/3a)+2a=3(x-2/3a)？ +2a-4/3a？

So the question becomes looking for 2a-4/3a? & gt0

a(2/3a- 1)& lt； 0

Solve this inequality and get 0.

I don't know if these questions are all right, but there is definitely no problem with the thinking. You can do the math yourself.

We can use this kind of thinking when solving this kind of problems in the future. As long as you know the derivative, the slope relationship between parallel lines and vertical lines, and the concept of tangent, you are not afraid to do these questions.