The table shows experimental values of two variables x and y. It is known that x and y are related by the equation $y = A k^x$, where $A$ and $k$ are constants.

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

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

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

Plot the points $(1, 0.420)$, $(2, 0.590)$, $(3, 0.750)$, $(4, 0.920)$, and $(5, 1.080)$ on the grid with $\lg y$ on the vertical axis and $x$ 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.

$y = A k^x$
Taking logarithms to base 10 on both sides:
$\lg y = \lg(A k^x)$
$\lg y = \lg A + x \lg k$
Comparing this to $Y = mX + c$, where $Y = \lg y$ and $X = x$:
Gradient $m = \lg k$
Vertical intercept $c = \lg A$

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

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

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

Using the point $(1, 0.420)$ and gradient $m = 0.165$:
$0.420 = 0.165(1) + c$
$c = 0.420 - 0.165 = 0.255$
$\lg A = 0.255$
$A = 10^{0.255} \approx 1.80$

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

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

2. Identify the variables to plot on the axes

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

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

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