下表给出了两个变量 $x$ 与 $y$ 的实验值。已知 $x$ 与 $y$ 通过形如 $y = h(x+1)^k$ 的方程相关联，其中 $h$ 和 $k$ 为常数。

$x$	1	2	3	4	5	6
$y$	5.2	7.2	9.1	10.9	12.6	14.2

1. Calculate the values of $X = \lg(1+x)$ and $Y = \lg y$ for each data point.

For $x=1, y=5.2$: $X = \lg(2) \approx 0.301$, $Y = \lg(5.2) \approx 0.716$
For $x=2, y=7.2$: $X = \lg(3) \approx 0.477$, $Y = \lg(7.2) \approx 0.857$
For $x=3, y=9.1$: $X = \lg(4) \approx 0.602$, $Y = \lg(9.1) \approx 0.959$
For $x=4, y=10.9$: $X = \lg(5) \approx 0.699$, $Y = \lg(10.9) \approx 1.037$
For $x=5, y=12.6$: $X = \lg(6) \approx 0.778$, $Y = \lg(12.6) \approx 1.100$
For $x=6, y=14.2$: $X = \lg(7) \approx 0.845$, $Y = \lg(14.2) \approx 1.152$

2. Plot the points on the grid and draw the line of best fit.

Plot the points $(0.301, 0.716)$, $(0.477, 0.857)$, $(0.602, 0.959)$, $(0.699, 1.037)$, $(0.778, 1.100)$, $(0.845, 1.152)$ on axes with scale 2 cm to 0.1 unit.
Draw a straight line of best fit passing through or close to these points.

1. Convert the non-linear equation to linear form using logarithms

$y = h(x+1)^k$
$\lg y = \lg(h(x+1)^k)$
$\lg y = \lg h + k \lg(x+1)$

2. Identify the gradient and $Y$-intercept from the linear form

Comparing $\lg y = k \lg(x+1) + \lg h$ with $Y = mX + c$:
Gradient $m = k$
$Y$-intercept $c = \lg h$

3. Calculate the gradient to find $k$

Using points from the line of best fit, e.g., $(0.30, 0.72)$ and $(0.85, 1.00)$:
$k = \frac{1.00 - 0.72}{0.85 - 0.30} = \frac{0.28}{0.55} \approx 0.51$
(Acceptable range for $k$: $0.50$ to $0.54$)

4. Find the $Y$-intercept to calculate $h$

From the graph, the $Y$-intercept is approximately $0.56$.
$\lg h = 0.56$
$h = 10^{0.56} \approx 3.63$
(Acceptable range for $h$: $3.5$ to $3.8$)

1. Explain the process of finding the corresponding $y$ value from the graph

Calculate the value on the horizontal axis: $\lg(1+1.5) = \lg 2.5$.
Read the corresponding value on the vertical axis from the graph, which gives $\lg y$.
Calculate $y$ using $y = 10^{\lg y}$.