Can we fit a quadratic to this given data?

Question

$x\vphantom{4}$	$4$	$7$	$10$	$13$
$\smash{y}\vphantom{4}$	$96$	$133$	$160$	$117$

In the above table the values of $y$ were calculated from a formula $y = Ax^2 + Bx$ , where $A$ and $B$ are constants. One of the values of $y$ , however, has been deliberately misprinted.

Plot the values of $y/x$ against those of $x$ , and

find which value of $y$ has been misprinted and give the correct value,
estimate the numerical values of $A$ and $B$ .

Suggestion

In the above table the values of $y$ were calculated from a formula $y = Ax^2 + Bx$ , where $A$ and $B$ are constants. One of the values of $y$ , however, has been deliberately misprinted.

Plot the values of $y/x$ against those of $x$ , …

Why does the question ask us to use $y/x$ rather than $y$ ?

Solution

$x\vphantom{4}$	$4$	$7$	$10$	$13$
$\smash{y}\vphantom{4}$	$96$	$133$	$160$	$117$

In the above table the values of $y$ were calculated from a formula $y = Ax^2 + Bx$ , where $A$ and $B$ are constants. One of the values of $y$ , however, has been deliberately misprinted.

Plot the values of

$y/x$

against those of

$x$

…

We have the following calculated data:

$x\vphantom{4}$	$4$	$7$	$10$	$13$
$\smash{y/x}\vphantom{4}$	$24$	$19$	$16$	$9$

The data are plotted below.

A plot of y/x against x.

If $y = Ax^2 + Bx$ , then $y/x = Ax + B$ .

This is the equation of a straight line, so, given the information in the question, we should expect to find a line through three of the four points plotted above. This is in fact the case!

A plot of y/x against x, together with an added line.

From the plot above, the incorrect value of $y$ is the one corresponding to $x=10$ , namely $160$ .

Since the correct value of $y/x$ must lie on the line, this value is $14$ , and so the correct value of $y$ is $14\times10=140$ .

The equation of the blue line is $y/x = Ax + B$ , so we can find the value of $A$ by finding the gradient of the line and the value of $B$ by finding its $y$ -intercept.

The gradient of the line in the plot above is (from looking at the points where $x=4$ and $x=13$ ) $\frac{\Delta y}{\Delta x} = \frac{9 - 24}{13 - 4} = -\frac{15}{9} = -\frac{5}{3}$ so $A = -\dfrac{5}{3}$ . Taking the point with $x=4$ , $y/x=24$ , we have $24=-\frac{5}{3}\times 4+B$ so $B=\dfrac{92}{3}=30\frac{2}{3}$ .

If we were doing this problem using real-world data rather than calculated figures, we would not use our data points to find the equation of the line (and hence $A$ and $B$ , but would instead use points on the line itself. In this case, though, these are the same thing.