Given the curve y = 1 / 3x3 + 4 / 3, find the tangent equation of the curve passing through the point P (2,4) The tangent point is not P Suppose the tangent point Q (a, a & # 179 / / 3 + 4 / 3) Slope f '(a) = A & # 178; y-(a³/3+4/3)=a²(x-a) Over P 4-a³/3-4/3=a²(2-a) a³-3a²+4=0 (a+1)(a-2)²=0 A = 2 is p So here a = - 1, so x-y-2 = 0 Where the slope f '(a) = A & # 178; how does it come from?

Given the curve y = 1 / 3x3 + 4 / 3, find the tangent equation of the curve passing through the point P (2,4) The tangent point is not P Suppose the tangent point Q (a, a & # 179 / / 3 + 4 / 3) Slope f '(a) = A & # 178; y-(a³/3+4/3)=a²(x-a) Over P 4-a³/3-4/3=a²(2-a) a³-3a²+4=0 (a+1)(a-2)²=0 A = 2 is p So here a = - 1, so x-y-2 = 0 Where the slope f '(a) = A & # 178; how does it come from?

y=1/3x3+4/3
The derivative of Y '= x & # 178;,
So the slope at x = a is a & # 178;

Given the curve y = 1 / 3x2 + 4 / 3, find the tangent equation of the curve passing through the point P (2,4)
X is followed by the square

Note that "over a certain point..." Then this point may not be the tangent point
1. If P is a tangent point, then the tangent slope k = f '(2);
2. If the point P is not a tangent point, let the tangent point be q (m, n), then the tangent slope k = f '(m) obtained from the derivative is equal to the slope of the straight line PQ. Then another equation about M and N is obtained by using the point Q on the curve, and the tangent slope is obtained by calculating the tangent point coordinates

Given that P is a point on the curve y = 3x3-2, and the abscissa of P is 1, how to solve the tangent equation of the curve at point P

y=3x³-2
y'=9x²
The tangent slope at point P is 9 & ﹥ 8226; 1 & ﹥ 178; = 9
The coordinates of point P are (1,1)
Let the tangent equation be y = 9x + B
If 9 &; 1 + B = 1, then B = - 8
The tangent equation at point P is y = 3x-8

Given the curve C: y = x 3 - x + 2 and point a (1, 2), the tangent equation of point a is obtained

∵ y = x3-x + 2, ∵ y ′ = 3x2-1, if point a (1,2) is tangent point, then the equation of k = 2 ∵ tangent is Y-2 = 2 (x-1), that is, 2x-y = 0. If point a is not tangent point, then let the tangent point be (x1, Y1), then Y1 = x13-x1 + 2, 3x12-1 = Y1 − 2x1 − 1, the solution is X1 = - 12, ∵ tangent equation is Y-2 = - 14 (x-1), that is