题目
Problem
Figure 6 shows a sketch of part of the curve with equation y=f(x), where
f(x)=(2x+5)(x−3)2.
Figure 6
(a) Deduce the values of x for which f(x)≤0.
(2)
The curve crosses the y-axis at the point P, as shown.
(b) Expand f(x) to the form
ax3+bx2+cx+d
where a, b, c and d are integers to be found.
(3)
(c) Hence, or otherwise, find
(i) the coordinates of P,
(ii) the gradient of the curve at P.
(2)
The curve with equation y=f(x) is translated two units in the positive x direction to a curve with equation y=g(x).
(d) (i) Find g(x), giving your answer in a simplified factorised form.
(ii) Hence state the y intercept of the curve with equation y=g(x).
(3)
解答
(a)
解法一
思路
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由于 (x−3)2≥0,函数符号主要由 2x+5 决定;但 x=3 时平方因式为 0,也要包含。
答题过程
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f(x)=(2x+5)(x−3)2.
Since (x−3)2≥0,
f(x)≤0
when
2x+5≤0
or when (x−3)2=0.
Therefore
x≤−25orx=3.
(b)
解法一
思路
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先展开平方,再乘以 (2x+5),最后合并同类项。
答题过程
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f(x)====(2x+5)(x−3)2(2x+5)(x2−6x+9)2x3−12x2+18x+5x2−30x+452x3−7x2−12x+45.
(c)
解法一
思路
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P 是 y 轴截距,所以令 x=0。斜率则用导函数在 x=0 的值。
答题过程
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At x=0,
f(0)=45.
Thus
P=(0,45).
Differentiate:
f′(x)=6x2−14x−12.
At P, x=0, so the gradient is
f′(0)=−12.
(d)
解法一
思路
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向右平移 2 个单位,要把 f(x) 中的 x 换成 x−2。然后令 x=0 求新曲线的 y 轴截距。
答题过程
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g(x)===f(x−2)(2(x−2)+5)(x−2−3)2(2x+1)(x−5)2.
At x=0,
g(0)=(1)(−5)2=25.
So the y intercept is 25, or (0,25).