The linear equation which passes through the point (- 1, - 2) and is tangent to the curve y = 2x-x ^ 3 is solved by derivative

The linear equation which passes through the point (- 1, - 2) and is tangent to the curve y = 2x-x ^ 3 is solved by derivative

Let the tangent point be (m, 2m-m & # 179;)
y'=2-3x²
k=2-3m²
The tangent equation is y = (2-3m & # 178;) (x-m) + 2m-m & # 179;
Substitute (- 1, - 2) to get M & # 178; (2m + 3) = 0
So m = 0 or - 3 / 2
The tangent equation is y = 2x or y = - 19x / 4-27 / 4

Given the curve C: y = x ^ 3-3x ^ 2 + 2x, the straight line y = KX, and l and C tangent to the point (x0, Y0) (x ≠ 0), the equation and tangent point coordinates of the straight line L are obtained
ditto.

Line L is tangent to curve C at point (x0, Y0)
=>
Function values are equal:
x0^3 - 3x0^2 +2x0 =kx0
=>
x0^2 - 3x0 + 2 =k
And the slopes are equal
3x0^2 - 6x0 +2 =k
=>
x0=3/2
=>
k=-1/4
y0=-3/8

The linear equation which is tangent to the curve y = 2x-x ^ 3 and parallel to the line y = x is? (it is required to use the knowledge of derivative to solve)

This is very simple
Take the derivative of y = 2x - x ^ 3 and get y '= 2 - 3x ^ 2. This is the slope
To be parallel to y = x, the slope is required to be 1
So: 2-3x ^ 2 = 1
We get x = 1 / root 3
Substituting into the equation of the curve, two tangent points are obtained: (1 / √ 3,5 / 3 √ 3) and (- 1 / √ 3, - 5 / 3 √ 3)
Substituting tangent point into straight line: y = x + B
Find B

Solving the linear equation of point (1, - 3) and tangent to curve y = x ^ 2

y=x^2
y'=2x
Let the tangent point be (a, a ^ 2), then the tangent is y = 2A (x-a) + A ^ 2 = 2ax-a ^ 2
Substitution point (1, - 3), - 3 = 2a-a ^ 2
That is, a ^ 2-2a-3 = 0
(a-3)(a+1)=0
a=3,-1
So there are two straight lines
y=6x-9
Or y = - 2x-1

If the line L is perpendicular to the line 2x-6y + 1 = 0 and tangent to the curve y = x ^ 3 + 3x ^ 2-1, then the equation function f (x) = the derivative of x ^ 2 + BX + C

Given that the line L is perpendicular to the line 2x-6y + 1 = 0, then the slope of the line L is - 3, the line L curve is y = x ^ 3 + 3x ^ 2-1, so the tangent point is when the derivative of Y is 3x ^ 2 + 6x = - 3, and the solution is x = - 1, y = 1, so the equation of the line L is y = - 3x-2, and the equation function of the line L is f (x) = x ^ 2 + BX + C

If all the images of parabola C1: y = 2kx2 + 3x + 1 are above the x-axis, and all the images of parabola C2: y = - x2 + 2x + 3k-7 are below the x-axis, try to find the value range of real number K
The answer is 9 / 8

The value of discriminant is less than 0
9-8k＜0,——（1）
4+4（3k-7）＜0——（2）
Solution (1) is k > 9 / 8, solution (2) is k < 2
Therefore, 9 / 8 < K < 2

Given that the parabola C1: y = 2x ^ 2 and the parabola C2 are symmetric with respect to the straight line y = - x, then the Quasilinear equation of C2?

Substituting y = - X and x = - y into y = 2x & # 178;
Then C2: x = - 2Y & # 178;
y²=-1/2x
The focus of C2 is on the negative half axis of Y axis
P=1/4
Then the alignment x = P / 2 = 1 / 8
reference resources

If the parabola C1: y = x2 + 1 and the parabola C2 are symmetric about the X axis, then the analytical expression of the parabola C2 is ()
A. y=-x2B. y=-x2+1C. y=x2-1D. y=-x2-1

On the change of the opening direction of the two analytic expressions of x-axis symmetric functions, the opening degree remains unchanged, and the coefficients of quadratic terms are opposite to each other; on the contrary of the intersection with y-axis, the constant terms are opposite to each other, so D

We know the function y = Y1 + Y2, where Y1 is positively proportional to x, Y2 is inversely proportional to X-2, and when x = 1, y = - 1, when x = 4, y = 13, the functional relation of this function is obtained

Let Y1 = K1X, y2 = K2 / (X-2) (K1, K2 are not equal to 0)
Then y = K1X + K2 / (X-2) is substituted into the data to obtain:
-1=k1-k2
13=4k1+k2/2
The solution is: K1 = 25 / 9, K2 = 34 / 9
So the function relation is: y = (25 / 9) x + (34 / 9) / (X-2)

Known function y = Y1 + Y2
Given the function y = Y1 + Y2, Y1 is proportional to x, Y2 is inversely proportional to x, and y = 4 when x = 1, y = 5 when x = 2
1 find the function relation between Y and X
2 the value of y when x = - 2

Let y = Y1 + Y2, where Y1 = K1X, y2 = K2 / X
Substituting: y = K1X + K2 / x,
Namely: 4 = K1 + K2 (1)
5=2k1+k2/2（2）
10=4k1+k2
(2) (1) we get 6 = 3k1, K1 = 2,
k2=2,
1.y=2x+2/x.
2. When x = - 2, y = 2 × (- 2) + 2 / (- 2) = - 5