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面积 x=-2y^2-4y+1,x=y^2+8y+10

解答

面积

x = - 2 y^{2} - 4 y + 1, x = y^{2} + 8 y + 10

解答

无解

求解步骤

对于

x = - 2 y^{2} - 4 y + 1

将

y

移到一边:

y = - \frac{4 + - 8 x + 24}{4}, y = - \frac{4 - - 8 x + 24}{4}

对于

x = y^{2} + 8 y + 10

将

y

移到一边:

y = - 4 + x + 6, y = - x + 6 - 4

f_{1} (x) = - \frac{4 + - 8 x + 24}{4}

f_{2} (x) = - \frac{4 - - 8 x + 24}{4}

要找到相交点，解

f_{i} (x) = f_{j} (x)

- \frac{4 + - 8 x + 24}{4} = - \frac{4 - - 8 x + 24}{4} : x = 3

- \frac{4 + - 8 x + 24}{4} = - 4 + x + 6 : x = 3, x = - 5

- \frac{4 + - 8 x + 24}{4} = - x + 6 - 4 : x \in R

无解

- \frac{4 - - 8 x + 24}{4} = - 4 + x + 6 : x = 3

- \frac{4 - - 8 x + 24}{4} = - x + 6 - 4 : x \in R

无解

- 4 + x + 6 = - x + 6 - 4 : x = - 6

曲线没有界限

= 无解

作图

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