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IAL 2021 June Q8

A Level / Edexcel / P1

IAL 2021 June Paper · Question 8

题目

Problem

The curve C1C_1 has equation

y=3x2+6x+9.\begin{align*} y=3x^2+6x+9. \end{align*}

(a) Write 3x2+6x+93x^2+6x+9 in the form

a(x+b)2+c\begin{align*} a(x+b)^2+c \end{align*}

where aa, bb and cc are constants to be found.

(3)

The point PP is the minimum point of C1C_1.

(b) Deduce the coordinates of PP.

(1)

A different curve C2C_2 has equation

y=Ax3+Bx2+Cx+D\begin{align*} y=Ax^3+Bx^2+Cx+D \end{align*}

where AA, BB, CC and DD are constants.

Given that C2C_2

  • passes through PP
  • intersects the xx-axis at 4-4, 2-2 and 33

(c) find, making your method clear, the values of AA, BB, CC and DD.

(5)

解答

(a)

解法一

思路

展开

先提出 33,再配方。

答题过程

展开 3x2+6x+9=3(x2+2x)+9=3((x+1)21)+9=3(x+1)2+6.\begin{align*} 3x^2+6x+9 =&\,3(x^2+2x)+9\\ =&\,3\bigl((x+1)^2-1\bigr)+9\\ =&\,3(x+1)^2+6. \end{align*}

(b)

解法一

思路

展开

配方形式 3(x+1)2+63(x+1)^2+6 中,平方项最小为 00,所以最小点在 x=1x=-1,最小值为 66

答题过程

展开

The minimum point is

P=(1,6).\begin{align*} P=(-1,6). \end{align*}

(c)

解法一

思路

展开

已知三次曲线的三个 xx 轴交点,所以可先写成因式形式。再用点 P(1,6)P(-1,6) 求整体倍数,最后展开得到 A,B,C,DA,B,C,D

答题过程

展开

Since the xx-intercepts are 4-4, 2-2 and 33,

y=α(x+4)(x+2)(x3).\begin{align*} y=\alpha(x+4)(x+2)(x-3). \end{align*}

The curve passes through P(1,6)P(-1,6), so

6=α(1+4)(1+2)(13)6=α(3)(1)(4)6=12αα=12.\begin{align*} 6=&\,\alpha(-1+4)(-1+2)(-1-3)\\ 6=&\,\alpha(3)(1)(-4)\\ 6=&\,-12\alpha\\ \alpha=&\,-\frac12. \end{align*}

Therefore

y=12(x+4)(x+2)(x3).\begin{align*} y=&\,-\frac12(x+4)(x+2)(x-3). \end{align*}

Expand:

(x+4)(x+2)(x3)=(x2+6x+8)(x3)=x33x2+6x218x+8x24=x3+3x210x24.\begin{align*} (x+4)(x+2)(x-3) =&\,(x^2+6x+8)(x-3)\\ =&\,x^3-3x^2+6x^2-18x+8x-24\\ =&\,x^3+3x^2-10x-24. \end{align*}

Hence

y=12x332x2+5x+12.\begin{align*} y =&\,-\frac12x^3-\frac32x^2+5x+12. \end{align*}

So

A=12,B=32,C=5,D=12.\begin{align*} A=-\frac12,\quad B=-\frac32,\quad C=5,\quad D=12. \end{align*}