Given the expression I(R+a) = b, linearize it to plot a suitable graph, identify the variables for the axes, and determine the expressions for the gradient and y-intercept.

The given expression is $I(R+a) = b$.
Expand the expression:

Rearrange the equation to match the linear form $Y = mX + C$. We want to plot $R$ against $\frac{1}{I}$.

Divide by $I$:

This can be written as:

Comparing this to $Y = mX + C$:
$Y = R$
$X = \frac{1}{I}$
$m = b$ (gradient)
$C = -a$ (y-intercept)

Calculate the reciprocal of current, $\frac{1}{I}$, for each given resistance $R$.

Plot a graph of $R$ (y-axis) against $\frac{1}{I}$ (x-axis). The graph should be a straight line.
The gradient of this straight line will give the value of $b$.
The y-intercept (the value of $R$ when $\frac{1}{I} = 0$) will give the value of $-a$.

The gradient $m = b$. Using the first point $(X_1, Y_1) = (5.00, 10)$ and the last point $(X_2, Y_2) = (25.00, 130)$ from the table for the line of best fit:

Therefore, $b = 6.0\,\text{V}$.

The y-intercept $C = -a$. Using the equation $Y = mX + C$, which is $R = b\left(\frac{1}{I}\right) - a$, and one point, for example $(X, Y) = (5.00, 10)$:

The values are: