How do you find the intercepts of #4x-3y=12#?

1 Answer
Mar 17, 2016

Answer:

#x# intercept: #(3,0)#
#y# intercept #(0;-4)#

Explanation:

You have to evaluate the equation for #x=0# and #y=0#

  1. #x_nn = x# intercept:
    #{(4x-3y=12),(y=0):}=>4x=12=>x=12/4=3=>P_(x_nn)(3;0)#

  2. #y_nn=y# intercept
    #{(4x-3y=12),(x=0):}=>-3y=12=>y=-12/3=-4=>P_(y_nn)(0;-4)#

graph{4x-3y-12=0 [-10, 10, -5, 5]}