How do I draw a straight line graph given a y=mx+c equation by the table method?

The y=mx+c equation is for a straight line (also known as 'linear') graph.

The 'y' is representative of the y-value, 'm' represents the gradient (how steep the graph is) and 'c' represents the y-intercept (where the graph cuts the y-axis).

To do this by the table method, lets take the following example:

Qu: Draw the graph of y=3x+4 for values of x between -3 and 3

We would draw a table as follows:

x:    -3   -2   -1   0   1   2    3      

3x:  -9   -6   -3   0   3   6    9

+4: +4  +4  +4  +4 +4  +4  +4

y:    -5   -2   -1   4   7  10   13

The first thing you do is write down the x-values for which the graph needs to be drawn (as specified in the qu, these are between -3 and 3)

You then put the 'mx' part of the equation in the next row. In our equation, m has the value of 3, so you write 3x (which algrebraically means 3 multiplied by x). Therefore, you multiple each x-value by 3, e.g -3x3=-9, -2x3=-6 etc. and write these values in.

You then put the +c part of the equation in the row below. In our equation, c has the value of 4, so you put +4 along the whole row. 

You then work out the y-value by adding the 2 rows above. Eg. For the value of x=-3, y=-9+4 which is equal to -5. Therefore, when x is -3, y is -5 - the coordinates would be (-3, -5). You do this for every value, which will give you a series of coordinates. 

You can then plot these coordinates on a graph and join them up to form your line graph!

 

NC

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