3.2 Understanding how equations are represented on a graph  (Page 3/4)

(b) y = – x + 3

• The y -intercept is 3, marked by a circle. The gradient is –1, which we change to $-\frac{1}{1}$ . This tells us to move one unit (denominator) to the right and then one unit (numerator) down (not up). We end up at the point ( 1 ; 2 ), marked by a second circle. Draw the line through the two circled points.

• Draw the following graphs using the y -intercept/gradient method you have just studied.

3.1 $y=-\mathrm{3x}+1$

3.2 $y=\frac{1}{3}x-\frac{5}{2}$

3.3 $y=\frac{-3}{4}x$

3.4 4 x – 3 y = 5

4 In problem 3.4 above, you should have written the equation in the standard form to be able to use m and c for the y –intercept/gradient method. This is a lot of extra work.

• There is another way to find the two points needed to be able to draw a straight–line graph. If we can find out where the graph cuts the x –axis as well as the y –axis, then we can simply draw the line through the two intercepts!

  	y – intercept	x – intercept
  y = 3 x – 4	( 0 ; –4 )	$\left(1\frac{1}{3};0\right)$
  y = –4 x + 3	( 0 ; 3 )	$\left(\frac{3}{4};0\right)$
  y = ½ x +1	( 0 ; 1 )	( –2 ; 0 )

Going back again to the previous six graph problems, the table shows the x– and y–intercepts for three of them in the form of coordinates.

The important thing to notice is that the y–intercept always has a zero in the position of the x–coordinate, and the x–intercept always has a zero in the position of the y–coordinate.

• This means that if we take the equation as it is and make the x zero and simplify to find the value of y , it will give us the y –intercept. Making the y zero and finding x , gives us the x –intercept. Here is how to do it for the equation 9 – 6 x = 3 y (definitely not in the standard form):

  

• Finding the y –intercept:

• Substitute 0 for x :

9 – 6(0) = 3 y 9 – 0 = 3 y 9 = 3 y 3 y = 9  y = 3

The y –intercept in coordinate form is ( 0 ; 3 )

• Mark this point on the graph.

• Finding the x –intercept:

• Substitute 0 for y :

9 – 6 x = 3(0) 9 – 6 x = 0 9 = 6 x 6 x = 9  $x=\frac{9}{6}=\frac{3}{2}$

The x –intercept in coordinate form is $\left(\frac{3}{2};0\right)$

• Mark this point on the graph. Finally draw a line through the two points. Alongside is the sketch.

• This is a very easy and convenient method. If you work carefully and with concentration, it won’t easily go wrong. Practise the method on the following equations:

4.1 4 y + 3 x = 4 4.2 6 y + 15 = 2 x 4.3 3 x + 4 y = 0

4.4 3 y + 5 = 4 x 4.5 2 y + 8 = 6 x 4.6 4 y – 2 x – 4 = 0

Does this method remind you of something?

5 We still have to consider a few special cases. With the equation written in the standard form, we can deduce a great deal about the graph

• The standard form of the straight–line equation is y = mx + c , as we know. If c is zero, then the equation becomes y = mx ; if m is zero, then the equation becomes y = c .

• 1 2 3 – 1 – 2– 3 0123– 1– 2– 3 y = mx + c , (where neither m nor c is zero), is the equation which produces lines which do not pass through the origin, nor are they horizontal or vertical. The first diagram shows some of these graphs. You can use either the y –intercept/gradient method or the two–intercept method for drawing these graphs.
• 1 2 3 – 1 – 2– 3 0123– 1– 2– 3 y = mx ( c is zero) produces lines which are neither horizontal nor vertical. They do pass through the origin, which is understandable as c is zero, meaning the y –intercept is zero. The second diagram shows a few of these. The y –intercept/gradient method is the simplest for drawing these graphs.
• y = c is the equation of a horizontal line, as you have already seen Draw them by drawing a horizontal line through the y –intercept ( c ).
• 1 2 3 – 1 – 2– 3 0123– 1– 2– 3If the equation of the line is x = k , where k is a constant, then this is a vertical line with k the x –intercept. Draw them by finding k on the x –axis and drawing a vertical line through that point.The third diagram shows some of the horizontal and vertical graphs.