图中显示了曲线 y=f(x)y = \text{f}(x)。该曲线穿过点 A(−1,0)A(-1, 0)、B(0,2)B(0, 2)、C(1,0)C(1, 0) 和 D(3,−3)D(3, -3)。 在不同的且标记清晰的图上,分别画出下列函数的草图:

  1. (a)

    y=−12f(x)y = -\frac{1}{2}\text{f}(x),

  2. (b)

    y=f(1−x)y = \text{f}(1-x)。

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完整解法

2(a)

1. Determine the new coordinates of the given points by multiplying their $y$-coordinates by $-\frac{1}{2}$.

The transformation y=−12f(x)y = -\frac{1}{2}\text{f}(x) affects only the yy-coordinates of the graph (a vertical scaling by factor 12\frac{1}{2} and a reflection in the xx-axis).
A(−1,0)→(−1,0)A(-1, 0) \rightarrow (-1, 0)
B(0,2)→(0,−1)B(0, 2) \rightarrow (0, -1)
C(1,0)→(1,0)C(1, 0) \rightarrow (1, 0)
D(3,−3)→(3,1.5)D(3, -3) \rightarrow (3, 1.5)

2. Determine the transformed horizontal asymptote and the correct shape of the curve.

The original horizontal asymptote is y=−1y = -1. Multiplying by −12-\frac{1}{2} gives the new asymptote:
y=−1×(−12)=0.5y = -1 \times \left(-\frac{1}{2}\right) = 0.5
Since the original curve approaches y=−1y = -1 from below (i.e. values are more negative than −1-1), the new curve will approach y=0.5y = 0.5 from above.
The local maximum BB becomes a local minimum, and the local minimum DD becomes a local maximum. Draw a smooth curve passing through these points with the specified features.

答案

A sketched curve passing through (−1,0)(-1, 0) and (1,0)(1, 0), with a local minimum at (0,−1)(0, -1), a local maximum at (3,1.5)(3, 1.5), and approaching the horizontal asymptote y=0.5y = 0.5 from above as xx gets large.

2(b)

1. Determine the transformation of coordinates.

Let a point on y=f(x)y = \text{f}(x) be (x,y)(x, y). For the new graph Y=f(1−X)Y = \text{f}(1-X), setting y=f(x)y = \text{f}(x) means x=1−X  ⟹  X=1−xx = 1 - X \implies X = 1 - x. The transformation maps every original point (x,y)(x, y) to (1−x,y)(1-x, y).

2. Calculate the new coordinates of the key points.

Apply X=1−xX = 1 - x to the given points:

  • A(−1,0)↦(1−(−1),0)=(2,0)A(-1, 0) \mapsto (1 - (-1), 0) = (2, 0)
  • Maximum B(0,2)↦(1−0,2)=(1,2)B(0, 2) \mapsto (1 - 0, 2) = (1, 2)
  • C(1,0)↦(1−1,0)=(0,0)C(1, 0) \mapsto (1 - 1, 0) = (0, 0)
  • Minimum D(3,−3)↦(1−3,−3)=(−2,−3)D(3, -3) \mapsto (1 - 3, -3) = (-2, -3)

3. Determine the new behavior of the asymptote.

The original graph has an asymptote y=−1y = -1 as x→∞x \to \infty. For the transformed graph, as x→∞x \to \infty, the new XX-coordinate becomes 1−x→−∞1 - x \to -\infty. Thus, the horizontal asymptote is y=−1y = -1 as x→−∞x \to -\infty.

答案

A correctly drawn sketch with the following features:

  • Passes through (0,0)(0, 0) and (2,0)(2, 0)
  • Local maximum at (1,2)(1, 2)
  • Local minimum at (−2,−3)(-2, -3)
  • Horizontal asymptote y=−1y = -1 approached from below as x→−∞x \to -\infty

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