题目
Problem
The curve C1 has equation
y=3x2+6x+9.
(a) Write 3x2+6x+9 in the form
a(x+b)2+c
where a, b and c are constants to be found.
(3)
The point P is the minimum point of C1.
(b) Deduce the coordinates of P.
(1)
A different curve C2 has equation
y=Ax3+Bx2+Cx+D
where A, B, C and D are constants.
Given that C2
- passes through P
- intersects the x-axis at −4, −2 and 3
(c) find, making your method clear, the values of A, B, C and D.
(5)
解答
(a)
解法一
思路
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先提出 3,再配方。
答题过程
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3x2+6x+9===3(x2+2x)+93((x+1)2−1)+93(x+1)2+6.
(b)
解法一
思路
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配方形式 3(x+1)2+6 中,平方项最小为 0,所以最小点在 x=−1,最小值为 6。
答题过程
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The minimum point is
P=(−1,6).
(c)
解法一
思路
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已知三次曲线的三个 x 轴交点,所以可先写成因式形式。再用点 P(−1,6) 求整体倍数,最后展开得到 A,B,C,D。
答题过程
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Since the x-intercepts are −4, −2 and 3,
y=α(x+4)(x+2)(x−3).
The curve passes through P(−1,6), so
6=6=6=α=α(−1+4)(−1+2)(−1−3)α(3)(1)(−4)−12α−21.
Therefore
y=−21(x+4)(x+2)(x−3).
Expand:
(x+4)(x+2)(x−3)===(x2+6x+8)(x−3)x3−3x2+6x2−18x+8x−24x3+3x2−10x−24.
Hence
y=−21x3−23x2+5x+12.
So
A=−21,B=−23,C=5,D=12.