将 −2x2+3x+2-2x^{2} + 3x + 2 表示为 a(x+b)2+ca(x + b)^{2} + c 的形式,其中 aa、bb 和 cc 是常数。 利用第 (a) 部分的答案,解释为什么方程为 y=−2x2+3x+2y = -2x^{2} + 3x + 2 和 (x−9)2+(y−15)2=25(x - 9)^{2} + (y - 15)^{2} = 25 的两条曲线不会相交。

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完整解法

1(a)

1. Factor out the coefficient of $x^2$ from the $x$ terms

−2x2+3x+2=−2(x2−32x)+2-2x^{2} + 3x + 2 = -2\left(x^{2} - \frac{3}{2}x\right) + 2

2. Complete the square for the expression inside the bracket

=−2[(x−34)2−(34)2]+2= -2\left[ \left(x - \frac{3}{4}\right)^{2} - \left(\frac{3}{4}\right)^{2} \right] + 2
=−2[(x−34)2−916]+2= -2\left[ \left(x - \frac{3}{4}\right)^{2} - \frac{9}{16} \right] + 2

3. Expand and simplify to reach the final form

=−2(x−34)2+98+2= -2\left(x - \frac{3}{4}\right)^{2} + \frac{9}{8} + 2
=−2(x−34)2+258= -2\left(x - \frac{3}{4}\right)^{2} + \frac{25}{8}

答案

−2(x−34)2+258-2\left(x - \frac{3}{4}\right)^{2} + \frac{25}{8}

1(b)

1. Determine the maximum $y$-value of the parabola using the result from part (a)

From part (a), y=−2x2+3x+2=−2(x−34)2+258y = -2x^2 + 3x + 2 = -2\left(x - \frac{3}{4}\right)^2 + \frac{25}{8}.
Since −2(x−34)2⩽0-2\left(x - \frac{3}{4}\right)^2 \leqslant 0 for all real xx, the maximum value of yy for the parabola is 258=3.125\frac{25}{8} = 3.125.

2. Identify the center and radius of the circle

The equation of the circle is (x−9)2+(y−15)2=25(x - 9)^2 + (y - 15)^2 = 25.
Comparing with (x−a)2+(y−b)2=r2(x - a)^2 + (y - b)^2 = r^2, the center is (9,15)(9, 15) and the radius is r=25=5r = \sqrt{25} = 5.

3. Determine the minimum $y$-value of the circle and conclude

The minimum yy-value of the circle is the yy-coordinate of the center minus the radius:
Minimum y=15−5=10y = 15 - 5 = 10.
Since the maximum yy-value of the parabola (3.1253.125) is strictly less than the minimum yy-value of the circle (1010), the highest point of the parabola lies below the lowest point of the circle. Hence, the two curves will not intersect.

答案

The maximum yy-value of the parabola is 3.1253.125, which is less than the minimum yy-value of the circle (1010). Therefore, the curves cannot intersect.

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