Modelling with Graphs (AQA Level 3 Mathematical Studies (Core Maths): Paper 2C: Graphical Techniques): Revision Note

Jodie is tracking the value of her new car to model its depreciation.

She collects the following data using online valuations for her car for the first 5 years. The values are rounded to the nearest hundred pounds.

(a) Plot a graph of this data and join the points with a smooth curve.

Answer:

Plot the values as accurately as you can and then join with a smooth curve

(b) Suggest two types of model which may fit this data.

Answer:

Consider the shape of the graph
It is not a straight line, so it is not a linear graph
It is decreasing either towards a stationary point, which suggests it could be quadratic
Or it could be continually decreasing, tending towards a lower limit, which suggests an exponential model

Quadratic or Exponential

(c) Jodie chooses to model the data with an equation: $V=120t^2+bt+c$.

Using the data in the table, find possible values for $b$ and $c$.

Answer:

The value of $c$ is the $y$-intercept, when $t=0$
This can be seen in the table and on the graph, as 20 000

$\mathit{c}\mathbf{=}\mathbf{20}\mathbf{000}$

The model is now

$V=120t^2+bt+20000$

Use one of the data points from the table to substitute into the equation
Using the last value in the table, $V=10700$ and $t=5$

$10700=120\left(5^2\right)+b\left(5\right)+20000$

Solve for $b$

$\begin{array}{rcl}-9300& =& 120\left(5^2\right)+5b\\ -9300& =& 3000+5b\\ -12300& =& 5b\end{array}$

$\mathit{b}\mathbf{=}\mathbf{-}\mathbf{2460}$

Note that selecting a different data point may result in a different answer for $b$, which is why the question asks for "possible" values - it is an approximate model, not an exact formula

(d) Suggest why this equation may not be suitable to model the value of the car.

Answer:

The suggested model is $\mathit{V}\mathbf{=}\mathbf{120}{\mathit{t}}^{\mathbf{2}}\mathbf{-}\mathbf{2460}\mathit{t}\mathbf{+}\mathbf{20}\mathbf{000}$ which is a positive quadratic.

The downward part of the quadratic will model the decrease in value well, but a positive quadratic will also have an upward curve, which will not model the decline in value accurately at all.

Using this positive quadratic model would predict that the car would start increasing in value, at an increasing rate, after the minimum value, which is highly likely to be incorrect.