Method for finding the tangent equation for the curve 20, and the method for finding the tangent equ

Method 1: Derivative.

y'=2x+2

k=2x=2

x=1,y'=4

So k=4x=1, y=0

So tangent coordinates (1,0).

Point oblique. 4x-y-4=0

Method 2: Set the tangent line to y=k(x-1).

Substituting yields x 2+(2-k)x-3+k=0

From the problem, equations have only one real root.

i.e. =k 2-4k+4+12-4k=k 2-8k+16=(k-4) 2=0

The tangent equation of k=4 is y=4x-4

Derivation. y'=2x+2

The derivative is the slope of the tangent.

x=1,y'=4

So tangent slope = 4

x=1，y=0

So tangent point (1,0).

So y-0=4(x-1).

4x-y-4=0

It can also be done with the methods of elementary mathematics.

Let the tangent line be y=k(x-1).

substitution. x^2+(2-k)x-3+k=0

The tangent has only one intersection point with the curve.

So. =k^2-4k+4+12-4k=k^2-8k+16=(k-4)^2=0

The solution gives k=4, so y=4x-4

y=x 2+2x-3 is derived for x.

y'=2x+2Substituting x=1 to get y'=4

And when x=1, y=0, the tangent equation is y-0=4 (x-1), i.e., y=4x-4

Problem 1: Find the tangent equation of the curve by the title:

Derivative of the function, which is the derivative of the composite function.

Let t=x+1, then the original function is composed of y=1 t **t=1+x.

y'=(1/t)'*1+x)'=1/t^2*1=-1/(1+x)^2

Let x=1,y'=-1/2^2=-1/4

So the slope of the tangent at point A is -1 4, so the tangent equation: y=-1 4 *(x-1)+1 2, i.e. y=-x 4+3 4

Problem 2: How to find the tangent equation of a curve Curve c: y=f(x), the point p(a, f(a)) on the curve

f(x) of the derivative f'(x) Existence.

1) Tangent equation with p as the tangent point: y-f(a)=f'(a)(x-a)

For example, the known function f(x) = (3x 2+6x-6) (x-1) finds the tangent equation of the function f(x) at the point (-1, 9 2);

f(x)=(3x^2+6x-6)/(x-1)=[3x^2-3x)+(9x-9)+3]/(x-1)=(3x+9)+3/(x-1)

f(-1)=(3-6-6) (1-1)=9 2, i.e., the point (-1,9 2) on the function image, f (x)=3-3 (x-1) 2, f (-1)=3-3 (-1-1) 2=9 4, so the tangent equation is y-9 2=(9 4)(x+1), i.e., y=(9 4)x+27 4

2) If there is a tangent of curve c through p, and the tangent point is q(b,f(b)), then the tangent is y-f(a)=f'(b)(x-a), also y-f(b)=f'(b)(x-b), and [f(b)-f(a)] b-a)=f'(b)

For example, find the tangent equation for hyperbolic y=1 x crossing point (1,0)).

For the hyperbola y=1 x, f(x)=1 x, and the derivative f(x)=-1 (x 2), because f(1)=1 1=1≠0, the point p(1,0) is not on this hyperbola.

Let the straight line through p(1,0) be tangent to the hyperbola at the point t(a,f(a)), then the slope of the tangent line is k=[f(a)-0] (a-1)=f (a)=-1 (a 2), i.e., (1 a) (a-1)=-1 (a 2), and the solution is a=0 (then f(a)=f(0) is not defined, rounded) or a=1 2

So the tangent equation is y-0=(1 2)(x-1).

i.e. x-2y-1=0

Tangent equation with p as the tangent point: y-f(a)=f'(a)(x-a)；If there is a tangent of curve c through p, and the tangent point is q(b,f(b)), then the tangent is y-f(a)=f'(b)(x-a), also y-f(b)=f'(b)(x-b), and [f(b)-f(a)] b-a)=f'(b)。

If a point is on a curve.

Let the curve equation be y=f(x) and a point on the curve be (a,f(a)).

Find the derivative of the curve equation and get f'(x), substituting a point to get f'(a), which is the tangent slope of the crossing point (a, f(a)), is obtained from the point oblique equation of the straight line. y-f(a)=f'(a)(x-a)

If a point is not on the curve.

Let the curve equation be y=f(x) and a point outside the curve is (a,b).

Find the derivative of the curve equation and get f'(x), let the tangent point be (x0,f(x0)), and substitute x0 into f'(x) to get the tangent slope f'(x0), from the point oblique equation of the straight line, the equation of the tangent line y-f(x0)=f'(x0)(x-x0), because (a,b) is on the tangent, substituting the obtained tangent equation, has: b-f(x0)=f'(x0)(a-x0), x0 is obtained, and the tangent equation obtained by substitution is obtained, that is, the tangent equation is obtained.

You need to know a point on the curve, and you can use the formula after you know it, and the formula is as follows:

Tangent equation with p as the tangent point: y-f(a)=f'(a)(x-a)

Basic information: Tangent equation is the study of tangent and the slope equation of tangent, involving geometry, algebra, physical vectors, quantum mechanics and other contents. is the study of the relationship between tangent coordinate vectors of geometric figures. There are vector methods and analytical methods.

Calculate the derivative f first'(x), the essence of the derivative is the slope of the curve, for example, there is a point ( on the function, and the derivative f of that point'(a)=c then shows that the tangent slope of the point k=c, assuming that this tangent equation is y=mx+n, then m=k=c, and ac+n=b, so y=cx+b-ac

Formula: The derivative value is taken as the slope k and then the original point (x0, y0) is used, and the tangent equation is (y-b)=k(x-a).

Example: Find the tangent equation for the curve y=x -2x at (-1,3).

Solution: The problem says at (-1,3), which means that the coordinate must be y=x -2x on the curve

y'=2x-2

Tangent slope = y'|(x=-1)=2(-1)-2=-4, so the tangent equation is y-3=-4(x+1).

That is, 4x+y+1=0

So the answer is 4x+y+1=0.

Ellipses have formulas.

For example, the ellipse is x 2 a 2+y 2 b 2=11then the tangent equation at the point above ( is.

x0)x/2+(y0)y/2=1

2.The point n that is not on the curve can also be based on the idea in 1.

Let mn tangent ellipse be n(x0,y0), where x0,y0 is unknown and the tangent equation of n(x0,y0) is established according to the 1 method, then m(x,y) brings the m coordinates into the straight line to obtain a one-time equation about x0,y0, and (x0,y0) on the ellipse, and satisfies the equation of the ellipse (2 times), and the two equations can be solved to solve two sets (x0,y0).

In fact, for any 2nd curve, the x 2 term in the curve equation can be changed to (x0)x, the y 2 term can be rewritten as (y0)y, x as x0, y as y0, x as x0, y as y0, and the constant term as unchanged to write the tangent equation at the (x0, y0) point on the curve.

Whether it is hyperbola, or parabola, or ellipse, or circle, it is applicable, and when the point is not on the curve, you can still use the idea in 2 above to find the tangent equation, so to speak, this is the general way to solve this kind of problem.