How to find a curve past a certain point, how to find the point at the inflection point of the curve

For example, (3m+4)x+(5-2m)y+7m-6=0 must pass a certain point to find this curve, no matter what real number m takes, and then find this fixed point.

3x-2y+7)m+(4x+5y-6)=0 This gives two equations. As long as 3x-2y+7 0, it doesn't matter what value M takes. In addition, if the equation is to be 0, then (4x+5y-6)=0 must also be true.

That is, no matter what value m takes, it will definitely pass the intersection of these two straight lines. The fixed point can be obtained by solving the following system of equations:

3x-2y+7＝0 （1）

4x+5y-6=0 （2）

1) 5 (2) 2.

15x+35+8x-12=0 x=-1

Substituting the solution yields y=2

So the equation you are talking about must pass through the point (-1,2).

Find the invariant relationship of the curve.

For example, y=kx 2 must pass (0,0).

Take your question as an example.

3m+4)x+(5-2m)y+7m-6=0, then (3m+4)*(x+1)-3m-4+(5-2m)(y-2)+10-4m+7m-6=0

Finishing: 3m+4)*(x+1)+(5-2m)(y-2)=0, that is, the curve must pass (-1,2)

The pump is set, the curve is also set, and there are many ways to get the pump to pass through this point, such as changing the speed, changing the impeller diameter, etc.

In the rebuttal method or the same method, a curve is set to pass a certain point, and then the opposite is demonstrated.

If the function y=f(x) is derivable at point c, and on one side of point c is convex and on the other side du is concave, then c is said to be the inflection point of the function y=f(x).

First, a derivative of the curve equation is performed.

The first derivative of y can be obtained as a unary quadratic polynomial.

Let it be zero, and solve the quadratic equation.

The solution is the inflection point.

Mathematically, it refers to the point at which the upward or downward direction of the curve changes, and intuitively the inflection point is the point at which the tangent crosses the curve (i.e., the point at which the curve is demarcated). If the function of the curve graph has a second derivative at the inflection point, the second derivative is either outlier (from positive to negative or negative to positive) or not present at the inflection point.

Order f''(x)=0, find x; Check f again''(x) on the left and right sides of x is not a different sign; If it is a different number, then.

x, f(x)) is the inflection point; If it is the same number, it is not an inflection point.

The specific grip is as follows:

1. According to the definition of the arubberization and throwing like a key object: the distance to the focal point is equal to the alignment, the distance from the point a to the alignment is 5, and then the abscissa of the point a is -4.

2. So the coordinates of the point p are (-4, 4), and then the distance between the two points can be obtained by using the distance formula.

This is a problem for the application of differential equations.

Solve the jujube coarseness: Suppose the normal equation of the curve is .

y-y0=-1/k(x-x0)

Rule. y-1=-x/k

y'=k=-x/(y-1)

Categorical variables, there are.

y-1)dy=-xdx

Points on both sides, get.

y-1)dy=∫-xdx

1/2·(y-1)²=1/2·x²+c

y-1)²=x²+2c

Rotten rock rot by starvation leakage over (0,1) point in the curve, so.

c=0 Therefore, the equation for the curve is .

x²+(y-1)²=0

Theorem 2: The sufficient and necessary condition for the curve system f(x,y,k)=0 with parameter k over the fixed point m(x0,y0) is that x=x0, y=y0 is always suitable for the equation f(x,y,k)=0, regardless of the value of k.

The basic steps to find a curve with one parameter are as follows:

1) Let the parameters take two different specific values to obtain two specific curve equations in the curve system;

2) the obtained two-curve equations are composed of equations to obtain the actual solution;

3) The obtained real number solution is substituted into the equation test of the original curve system, and if theorem 2 is satisfied, then this solution is the fixed point through which the curve system is sought.

Find the invariant relationship of the curve.

For example, y=kx 2 must pass (0,0).

Take your question as an example.

3m+4)x+(5-2m)y+7m-6=0, then (3m+4)*(x+1)-3m-4+(5-2m)(y-2)+10-4m+7m-6=0