题目
Problem
(a) Sketch the graph of the curve C with equation
y=x−k4,
where k is a positive constant.
Show on your sketch
- the coordinates of any points where C cuts the coordinate axes
- the equation of the vertical asymptote to C
(4)
Given that the straight line with equation y=9−x does not cross or touch C,
(b) find the range of values of k.
(5)
解答
(a)
解法一
思路
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这是 reciprocal graph。分母为 0 的位置给出竖直渐近线 x=k。由于 k>0,这条渐近线在 y 轴右侧。令 x=0 可得 y 轴截距;但分子恒为 4,所以不会有 x 轴截距。
答题过程
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The vertical asymptote occurs when the denominator is zero:
x−k=x=0k.
The y-intercept is found by setting x=0:
y=0−k4=−k4.
So the curve cuts the y-axis at
(0,−k4).
There is no x-intercept, since
x−k4=0
has no solution.
The sketch should show a decreasing reciprocal curve with vertical asymptote x=k, a right-hand branch above the x-axis, and a left-hand branch passing through (0,−k4).
(b)
解法一
思路
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直线不与曲线相交或相切,意思是联立后没有实根。联立得到关于 x 的二次方程,再令判别式小于 0。
答题过程
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At an intersection,
x−k4=9−x.
Multiplying by x−k,
4==(9−x)(x−k)9x−9k−x2+kx.
Hence
x2−(9+k)x+9k+4=0.
For the line not to cross or touch the curve, this quadratic must have no real roots, so
Δ[−(9+k)]2−4(1)(9k+4)(9+k)2−36k−16k2+18k+81−36k−16k2−18k+65(k−13)(k−5)<0<0<0<0<0<0.
Therefore
5<k<13.