How do you graph #2x-3y+15=0#?

1 Answer
Feb 7, 2017

Answer:

Use one of 3 methods.

Explanation:

There are 3 methods to plot a graph from a given equation:

  1. Plot points: Choose #x#-values and work out #y#-values
    It is good practice to choose five # x#-values. (2 negs, 0 and 2 pos)

  2. y-intercept /gradient method. Change the equation to #y = mx+c#
    Plot the y-intercept and use #m = "rise"/"run"# from that point.

  3. -The intercept method. Find both intercepts:
    To find the #x#-intercept, make #y =0#
    To find the #y#-intercept, make #x =0#

(This does not work if the line passes through the origin.)

#2x-3y+15 =0#

#2x+15 = 3y#

#y = 2/3x+5#

#1#. Choose x values which are divisible by 3

#x: -6" "-3" "0" "3" " 6" "larr# choose
#y: +1" "+3" "5" "7" "9" "larr# work out

Now plot the points and join with a straight line.

#2#. #y = 2/3x+5#

Plot the y-intercept at 5.
Count 2 up and 3 to the right, repeat.
Count 2 down and 3 to the left, repeat.

#3#. Use #2x -3y = -15#

x-int: #2x - 3(0) = -15 " " rarr x = -7 1/2#
y-int: #2(0) -3y =-15" "rarr y = 5#

Intercepts are at : #(-7 1/2, 0) and (0,5)#

The three methods will all give the graph shown:

graph{y = 2/3x+5 [-25.17, 14.83, -8.88, 11.12]}