题目
Problem
Figure 2 shows a sketch of the graph C1 with equation
y=−2x3+5x2+4x−3
and a sketch of the graph C2 with equation
y=a+∣5x+b∣
where a and b are constants.
Graphs C1 and C2 intersect at point P, point Q and point R, as shown in Figure 2.
Given that P has coordinates (−2,25),
(a) show that
a=15+b
(2)
Given also that R has coordinates (2,9),
(b) find the value of a and the value of b.
(3)
Using the answer to part (b),
(c) state the coordinates of the vertex of C2.
(2)
(d) Find, using algebra, the coordinates of Q. Show each stage of your working.
(Solutions relying on calculator technology are not acceptable.)
(6)
题目中文翻译
图 2 给出了图像 C1 的草图，其方程为
y=−2x3+5x2+4x−3
以及图像 C2 的草图，其方程为
y=a+∣5x+b∣
其中 a,b 为常数。
图像 C1 与 C2 相交于点 P,Q,R，如图所示。
已知 P 的坐标为 (−2,25)，
(a) 证明
a=15+b
又已知 R 的坐标为 (2,9)，
(b) 求 a,b 的值；
(c) 利用(b)的结果，写出 C2 顶点的坐标；
(d) 用代数方法求点 Q 的坐标，并写出每一步过程。
（不接受依赖计算器技术的解法。）
解答
(a)
At P(−2,25), the graph C2 gives
25=a+∣5(−2)+b∣
So
25=a+∣−10+b∣
From the sketch, P is on the left-hand branch of the absolute value graph, so 5x+b<0 at P.
Therefore
∣−10+b∣=−(−10+b)=10−b
Hence
25a=a+10−b=15+b
as required.
(b)
At R(2,9),
9=a+∣5(2)+b∣
From the sketch, R is on the right-hand branch, so 5x+b>0 at R.
Thus
9=a+10+b
Using a=15+b,
992bb=(15+b)+10+b=25+2b=−16=−8
Then
a=15+b=15−8=7
So
a=7,b=−8
(c)
Using part (b),
C2:y=7+∣5x−8∣
The vertex occurs when the expression inside the modulus is zero:
5x−8=0
So
x=58
At the vertex,
y=7
Therefore the vertex is
(58,7)
(d)
On the left-hand branch of C2,
5x−8<0
so
y=7−(5x−8)=15−5x
Point Q lies on this left-hand branch, so set this equal to C1:
15−5x=−2x3+5x2+4x−3
Rearrange:
02x3−5x2−9x+18=−2x3+5x2+9x−18=0
Since P(−2,25) is already one intersection point, x=−2 is a root. Therefore (x+2) is a factor.
Divide:
2x3−5x2−9x+18=(x+2)(2x2−9x+9)
So
(x+2)(2x2−9x+9)=0
The quadratic factor gives
2x2−9x+9=0
Factorise:
2x2−9x+9=(2x−3)(x−3)
So
x=23orx=3
The value x=3 is not on the left-hand branch 5x−8<0, so it is not point Q.
Thus
x=23
Substitute into y=15−5x:
y=15−5(23)=15−215=215
Therefore
Q(23,215) 