How do you graph the line #4x+3y-12=0#?

1 Answer
Aug 4, 2016

#y# intercept of the function is present at the point: #(0,4)#
and #x# intercept of the function is present at the point: #(3,0)#
See full explanation for more detail:

Explanation:

In order to clearly recognise the #x# and #y# intercepts of the function we may convert the given function to linear gradient form. That being:
#y=mx+c#
Where:

#c# is the constant determining the #y# intercept.
and #m# is the gradient of the function.

Therefore, let us do this:
#->3y=-4x+12#
Dividing both sides by #3#, we get:
#y=-4/3x+4#

This implies that the #y# intercept of the function is 4.
We can prove this via the statement:
#y# intercepts where #x=0#:
Therefore:
#y=0*x+4#
This implies that the #y# intercept of the function is present at the point: #(0,4)#

We can thus determine the #x# intercept using the statement:
#x# intercepts where #y=0#:
#->0=-4/3x+3#
#:.4/3x=4#
#x=3#
This implies that the #x# intercept of the function is present at the point: #(3,0)#

If we plot these points on a Cartesian plane and draw a line between the two, the graph has been drawn.

Attached below is a graph of the function with the intercepts in frame:

graph{y=-4/3x+4 [-9.54, 10.46, -2.92, 7.08]}