Ontario 09 Academic (MPM1D)
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Finding the rule (Table of values or pattern)
Lesson

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

Equations of Lines!

We have:

$y=mx+b$y=mx+b  (slope intercept form)

$y-y_1=m\left(x-x_1\right)$yy1=m(xx1)   (point slope formula)

$\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$yy1xx1=y2y1x2x1   (two point formula)

$ax+by+c=0$ax+by+c=0   (general form)

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 slope 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 slope 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 slope 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 slope, we have 4 possible points to choose from and we can use the point slope formula to find the equation of the line. 

$y-y_1=m\left(x-x_1\right)$yy1=m(xx1)

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

$y-12$y12 $=$= $3\left(x-3\right)$3(x3)
$y-12$y12 $=$= $3x-9$3x9
$y$y $=$= $3x-9+12$3x9+12
$y$y $=$= $3x+3$3x+3

 

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 slope 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=y2y1x2x1

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

$m$m $=$= $\frac{y_2-y_1}{x_2-x_1}$y2y1x2x1
$m$m $=$= $\frac{0-25}{2-\left(-3\right)}$0252(3)
$m$m $=$= $\frac{-25}{5}$255
$m$m $=$= $-5$5

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

$m$m $=$= $\frac{y_2-y_1}{x_2-x_1}$y2y1x2x1
$m$m $=$= $\frac{-40-0}{10-2}$400102
$m$m $=$= $\frac{-40}{8}$408
$m$m $=$= $-5$5

 

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.

Slope 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 slope easier)

$m$m $=$= $\frac{y_2-y_1}{x_2-x_1}$y2y1x2x1
$m$m $=$= $\frac{-50-0}{12-2}$500122
$m$m $=$= $\frac{-50}{10}$5010
$m$m $=$= $-5$5

Got it! A linear relationship.  

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

$y-y_1$yy1 $=$= $m\left(x-x_1\right)$m(xx1)
$y-0$y0 $=$= $-5\left(x-2\right)$5(x2)
$y$y $=$= $-5x+10$5x+10

 

Let's have a look at these worked examples.

Question 1

Use the table of values below to write an equation for $g$g in terms of $f$f.

$f$f $4$4 $5$5 $6$6 $7$7 $8$8
$g$g $8$8 $10$10 $12$12 $14$14 $16$16

 

 

 

Outcomes

9D.LR3.01

Determine values of a linear relation by using a table of values, by using the equation of the relation, and by interpolating or extrapolating from the graph of the relation

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