曲线 CC 的方程为 y=2x2−10x+17x−3,x∈R,x≠3y = \frac{2x^2 - 10x + 17}{x - 3}, \quad x \in \mathbb{R}, \quad x \ne 3。

  1. (i)

    在不使用计算器的情况下,求 CC 可以取到的 yy 值集合。

  2. (ii)

    草拟 CC 的图像,并用精确形式标出相关特征。

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完整解法

6(i)

1. Rearrange the equation of the curve into a quadratic equation in $x$.

y=2x2−10x+17x−3y = \frac{2x^2 - 10x + 17}{x - 3}

y(x−3)=2x2−10x+17y(x - 3) = 2x^2 - 10x + 17

xy−3y=2x2−10x+17xy - 3y = 2x^2 - 10x + 17

2x2−(10+y)x+(17+3y)=02x^2 - (10 + y)x + (17 + 3y) = 0

2. Apply the discriminant condition for $x$ to have real roots and simplify into a quadratic inequality in $y$.

Since the curve exists for real values of xx, the quadratic equation must have real roots. Thus, the discriminant Δ≥0\Delta \ge 0.

[−(10+y)]2−4(2)(17+3y)≥0[-(10 + y)]^2 - 4(2)(17 + 3y) \ge 0

100+20y+y2−8(17+3y)≥0100 + 20y + y^2 - 8(17 + 3y) \ge 0

100+20y+y2−136−24y≥0100 + 20y + y^2 - 136 - 24y \ge 0

y2−4y−36≥0y^2 - 4y - 36 \ge 0

3. Find the critical values by solving the corresponding quadratic equation.

Solving y2−4y−36=0y^2 - 4y - 36 = 0 using the quadratic formula:

y=−(−4)±(−4)2−4(1)(−36)2(1)y = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-36)}}{2(1)}

y=4±16+1442y = \frac{4 \pm \sqrt{16 + 144}}{2}

y=4±1602y = \frac{4 \pm \sqrt{160}}{2}

y=4±4102y = \frac{4 \pm 4\sqrt{10}}{2}

y=2±210y = 2 \pm 2\sqrt{10}

4. Determine the solution intervals for the quadratic inequality.

The quadratic curve for y2−4y−36≥0y^2 - 4y - 36 \ge 0 is a U-shaped parabola. The inequality is satisfied for values outside the roots.

y≤2−210ory≥2+210y \le 2 - 2\sqrt{10} \quad \text{or} \quad y \ge 2 + 2\sqrt{10}

答案

y≤2−210y \le 2 - 2\sqrt{10} or y≥2+210y \ge 2 + 2\sqrt{10}

6(ii)

1. Rewrite the function to find its asymptotes.

By algebraic manipulation or algebraic long division:
y=2x2−10x+17x−3=2x(x−3)−4x+17x−3y = \frac{2x^2 - 10x + 17}{x - 3} = \frac{2x(x - 3) - 4x + 17}{x - 3}
y=2x(x−3)−4(x−3)+5x−3=2x−4+5x−3y = \frac{2x(x - 3) - 4(x - 3) + 5}{x - 3} = 2x - 4 + \frac{5}{x - 3}
As x→±∞x \to \pm\infty, y→2x−4y \to 2x - 4. As x→3x \to 3, y→±∞y \to \pm\infty.
Therefore, the vertical asymptote is x=3x = 3 and the oblique asymptote is y=2x−4y = 2x - 4.

2. Find the exact coordinates of the turning points.

Differentiating with respect to xx:
dydx=2−5(x−3)2\frac{\mathrm{d}y}{\mathrm{d}x} = 2 - \frac{5}{(x - 3)^2}
At the stationary points, dydx=0\frac{\mathrm{d}y}{\mathrm{d}x} = 0:
5(x−3)2=2  ⟹  (x−3)2=52\frac{5}{(x - 3)^2} = 2 \implies (x - 3)^2 = \frac{5}{2}
x−3=±52=±102  ⟹  x=3±102x - 3 = \pm\sqrt{\frac{5}{2}} = \pm\frac{\sqrt{10}}{2} \implies x = 3 \pm \frac{\sqrt{10}}{2}
From part (i), or by direct substitution, the corresponding yy-values are y=2±210y = 2 \pm 2\sqrt{10}.
The minimum point is (3+102,2+210)\left(3 + \frac{\sqrt{10}}{2}, 2 + 2\sqrt{10}\right) and the maximum point is (3−102,2−210)\left(3 - \frac{\sqrt{10}}{2}, 2 - 2\sqrt{10}\right).

3. Find the axial intercepts.

When x=0x = 0:
y=17−3=−173y = \frac{17}{-3} = -\frac{17}{3}
The yy-intercept is (0,−173)\left(0, -\frac{17}{3}\right).
To find xx-intercepts, set 2x2−10x+17=02x^2 - 10x + 17 = 0. The discriminant is (−10)2−4(2)(17)=−36<0(-10)^2 - 4(2)(17) = -36 < 0, so there are no xx-intercepts.

4. Sketch the curve with all required features accurately positioned and labelled.

Draw axes and the asymptotes (x=3x=3 and y=2x−4y=2x-4). Sketch the left branch passing through (0,−173)\left(0, -\frac{17}{3}\right), turning at the maximum point (3−102,2−210)\left(3 - \frac{\sqrt{10}}{2}, 2 - 2\sqrt{10}\right) and approaching x=3x=3 downwards. Sketch the right branch coming from x=3x=3 downwards, turning at the minimum point (3+102,2+210)\left(3 + \frac{\sqrt{10}}{2}, 2 + 2\sqrt{10}\right), and going upwards along y=2x−4y=2x-4.

答案

Features to label on the sketch:

  • Vertical asymptote: x=3x = 3
  • Oblique asymptote: y=2x−4y = 2x - 4
  • yy-intercept: (0,−173)\left(0, -\frac{17}{3}\right)
  • Local minimum: (3+102,2+210)\left(3 + \frac{\sqrt{10}}{2}, 2 + 2\sqrt{10}\right)
  • Local maximum: (3−102,2−210)\left(3 - \frac{\sqrt{10}}{2}, 2 - 2\sqrt{10}\right)

(The curve consists of two branches bounded by the asymptotes. The local maximum is in the bottom-left region, crossing the yy-axis, and the local minimum is in the top-right region.)

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