下表显示了两个变量 x 和 y 的实验值。已知 x 和 y 满足关系式 y=Akxy = A k^x,其中 AA 和 kk 是常数。

  1. (a)

    在下方的网格中,以 x 为横轴,lg y 为纵轴进行描点,并画出一条直线以展示这些信息。

    (ii) 利用你的图像估计 AA 和 kk 的值。

  2. (b)

    变量 xx 和 yy 满足公式 y=ax3+bx2y = ax^3 + bx^2,其中 aa 和 bb 为常数。请清晰地解释如何通过画出一条直线来表示该公式。你应该说明各轴上应该绘制哪些变量,并解释如何计算出 aa 和 bb 的值。

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完整解法

12(a)

1. Calculate the values of $\lg y$ for each given $x$.

For x=1,2,3,4,5x = 1, 2, 3, 4, 5, the corresponding yy values are 2.63,3.89,5.62,8.32,12.022.63, 3.89, 5.62, 8.32, 12.02.
Calculating lg⁡y\lg y to 3 decimal places:
x=1  ⟹  lg⁡2.63≈0.420x = 1 \implies \lg 2.63 \approx 0.420
x=2  ⟹  lg⁡3.89≈0.590x = 2 \implies \lg 3.89 \approx 0.590
x=3  ⟹  lg⁡5.62≈0.750x = 3 \implies \lg 5.62 \approx 0.750
x=4  ⟹  lg⁡8.32≈0.920x = 4 \implies \lg 8.32 \approx 0.920
x=5  ⟹  lg⁡12.02≈1.080x = 5 \implies \lg 12.02 \approx 1.080

2. Plot the points and draw the straight line of best fit.

Plot the points (1,0.420)(1, 0.420), (2,0.590)(2, 0.590), (3,0.750)(3, 0.750), (4,0.920)(4, 0.920), and (5,1.080)(5, 1.080) on the grid with lg⁡y\lg y on the vertical axis and xx on the horizontal axis. Draw a straight line passing through these points.

3. Express the given equation in linear form to identify the gradient and vertical intercept.

y=Akxy = A k^x
Taking logarithms to base 10 on both sides:
lg⁡y=lg⁡(Akx)\lg y = \lg(A k^x)
lg⁡y=lg⁡A+xlg⁡k\lg y = \lg A + x \lg k
Comparing this to Y=mX+cY = mX + c, where Y=lg⁡yY = \lg y and X=xX = x:
Gradient m=lg⁡km = \lg k
Vertical intercept c=lg⁡Ac = \lg A

4. Calculate the gradient from the graph to find $k$.

Using points (1,0.420)(1, 0.420) and (5,1.080)(5, 1.080) from the line:
Gradient m=1.080−0.4205−1=0.6604=0.165m = \frac{1.080 - 0.420}{5 - 1} = \frac{0.660}{4} = 0.165
lg⁡k=0.165\lg k = 0.165
k=100.165≈1.46k = 10^{0.165} \approx 1.46

5. Calculate the vertical intercept from the graph to find $A$.

Using the point (1,0.420)(1, 0.420) and gradient m=0.165m = 0.165:
0.420=0.165(1)+c0.420 = 0.165(1) + c
c=0.420−0.165=0.255c = 0.420 - 0.165 = 0.255
lg⁡A=0.255\lg A = 0.255
A=100.255≈1.80A = 10^{0.255} \approx 1.80

答案

Graph plotted; A≈1.80A \approx 1.80, k≈1.46k \approx 1.46

12(b)

1. Transform the non-linear equation into a linear form $Y = mX + c$

Given y=ax3+bx2y = ax^3 + bx^2.
Divide both sides by x2x^2:
yx2=ax+b\frac{y}{x^2} = ax + b

2. Identify the variables to plot on the axes

Comparing yx2=ax+b\frac{y}{x^2} = ax + b with Y=mX+cY = mX + c, we have Y=yx2Y = \frac{y}{x^2} and X=xX = x.
Therefore, plot yx2\frac{y}{x^2} on the vertical axis and xx on the horizontal axis.

3. Explain how to calculate the constants $a$ and $b$

From the linear form, the gradient m=am = a and the YY-intercept c=bc = b.
Thus, aa can be calculated by finding the gradient of the straight line graph, and bb can be found by reading the vertical intercept.

答案

Plot yx2\frac{y}{x^2} on the vertical axis against xx on the horizontal axis. The value of aa is the gradient of the straight line, and the value of bb is the vertical intercept.

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