Finding the Rule II

So far we have four different ways to form the equation of a straight line.  

Question: If we are given a table of values, how can we be sure it forms a straight line? And how can we find the equation of that line?   

We can't just assume the points in a table of values are linear (make a straight line). So when given a table of values, we must first confirm that the values in the table form a linear relationship.

You can do this a couple of ways, depending on what sort of information is provided to you in the table.

Table has consecutive x values

What is it that makes a straight line? It is the fact that for each 1 unit change in $x$x, the $y$y value changes by the same amount each time (i.e. the gradient is always the same).

If the table shows you values of $y$y, for consecutive values of $x$x (like this one) then it is quite easy to both check for a linear relationship and find the gradient of that relationship.

Some things to note about this table: the $x$x values go up by 1 each time and it doesn't matter what number the table starts at.  

Notice in the above table, that for each 1 unit increase in $x$x, $y$y increases by $3$3 each time. This is a linear relationship, and what's more, this constant change in $y$y is the gradient of the line.

So we can check for a linear relationship by looking for a common difference between the $y$y values.  If each successive $y$y value has the same difference then it is linear. 

From here we not only have the gradient, we have 4 possible points to choose from and we can use the point gradient formula to find the equation of the line. 

$y-y_1=m\left(x-x_1\right)$y−y1​=m(x−x1​)

Using gradient = $3$3 and the point $\left(3,12\right)$(3,12):

Table has non-consecutive values of x

If the table shows you values of $y$y, for non-consecutive values of $x$x (like this one) then we need another way to check for a linear relationship.

Some things to note about this table: the $x$x values do not go up by 1 each time.

We can check for a linear relationship by either plotting accurately, and looking to see if it is a straight line, or by checking the gradient between all the points.  We will need to do this 3 times in this table.

$m=\frac{y_2-y_1}{x_2-x_1}$m=y2​−y1​x2​−x1​​

Gradient between $\left(-3,25\right)$(−3,25) and $\left(2,0\right)$(2,0):

Gradient between $\left(2,0\right)$(2,0) and $\left(10,-40\right)$(10,−40):

Looking good so far.  We now have shown that the $3$3 points $\left(2,0\right)$(2,0), $\left(10,-40\right)$(10,−40) and $\left(-3,25\right)$(−3,25) are collinear which means they are all on the one line.  We have to confirm that the last point is also on this line.

Gradient between $\left(2,0\right)$(2,0) and $\left(12,-50\right)$(12,−50):  

(see how I used the $\left(2,0\right)$(2,0) point again, this is because 0 values make evaluating the gradient easier)

Got it! A linear relationship.  

Now we can find its equation.  We already have the gradient, $m=-5$m=−5.  So we can use the point gradient formula.

Let's have a look at these worked examples.

Question 1