一条三次曲线穿过点 (2,3) 和 (−3,−22)。若该曲线在 (1,−6) 处有一个驻点,求其方程。

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完整解法

1

1. Set up the general equation for a cubic curve and find its derivative.

Let the equation of the cubic curve be y=ax3+bx2+cx+dy = ax^3 + bx^2 + cx + d.

Differentiating with respect to xx:
dydx=3ax2+2bx+c\frac{dy}{dx} = 3ax^2 + 2bx + c

2. Use the stationary point $(1, -6)$ to form two equations.

Since the point (1,−6)(1, -6) lies on the curve, we substitute x=1,y=−6x=1, y=-6 into the curve equation:
a(1)3+b(1)2+c(1)+d=−6a(1)^3 + b(1)^2 + c(1) + d = -6
a+b+c+d=−6— (1)a + b + c + d = -6 \quad \text{--- (1)}

Since (1,−6)(1, -6) is a stationary point, dydx=0\frac{dy}{dx} = 0 at x=1x = 1:
3a(1)2+2b(1)+c=03a(1)^2 + 2b(1) + c = 0
3a+2b+c=0— (2)3a + 2b + c = 0 \quad \text{--- (2)}

3. Use the points $(2, 3)$ and $(-3, -22)$ to form two more equations.

Substituting (2,3)(2, 3) into the curve's equation:
a(2)3+b(2)2+c(2)+d=3a(2)^3 + b(2)^2 + c(2) + d = 3
8a+4b+2c+d=3— (3)8a + 4b + 2c + d = 3 \quad \text{--- (3)}

Substituting (−3,−22)(-3, -22) into the curve's equation:
a(−3)3+b(−3)2+c(−3)+d=−22a(-3)^3 + b(-3)^2 + c(-3) + d = -22
−27a+9b−3c+d=−22— (4)-27a + 9b - 3c + d = -22 \quad \text{--- (4)}

4. Eliminate $d$ to form a system of equations in terms of $a$, $b$, and $c$.

From (1), make dd the subject:
d=−6−a−b−cd = -6 - a - b - c

Substitute dd into (3):
8a+4b+2c+(−6−a−b−c)=38a + 4b + 2c + (-6 - a - b - c) = 3
7a+3b+c=9— (5)7a + 3b + c = 9 \quad \text{--- (5)}

Substitute dd into (4):
−27a+9b−3c+(−6−a−b−c)=−22-27a + 9b - 3c + (-6 - a - b - c) = -22
−28a+8b−4c=−16-28a + 8b - 4c = -16
Divide by −4-4:
7a−2b+c=4— (6)7a - 2b + c = 4 \quad \text{--- (6)}

5. Solve the reduced system for the remaining constants.

Subtract (6) from (5) to eliminate aa and cc:
(7a+3b+c)−(7a−2b+c)=9−4(7a + 3b + c) - (7a - 2b + c) = 9 - 4
5b=5  ⟹  b=15b = 5 \implies b = 1

Substitute b=1b = 1 into (2) and (5):
From (2): 3a+2(1)+c=0  ⟹  c=−3a−2— (7)3a + 2(1) + c = 0 \implies c = -3a - 2 \quad \text{--- (7)}
From (5): 7a+3(1)+c=9  ⟹  c=6−7a— (8)7a + 3(1) + c = 9 \implies c = 6 - 7a \quad \text{--- (8)}

Equating (7) and (8):
−3a−2=6−7a-3a - 2 = 6 - 7a
4a=8  ⟹  a=24a = 8 \implies a = 2

Substitute a=2a = 2 back into (7):
c=−3(2)−2=−8c = -3(2) - 2 = -8

Substitute a,b,a, b, and cc into the expression for dd:
d=−6−2−1−(−8)=−1d = -6 - 2 - 1 - (-8) = -1

6. Write the final equation of the cubic curve.

With a=2a = 2, b=1b = 1, c=−8c = -8, and d=−1d = -1, the equation of the cubic curve is:
y=2x3+x2−8x−1y = 2x^3 + x^2 - 8x - 1

答案

y=2x3+x2−8x−1y = 2x^3 + x^2 - 8x - 1

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