What are the #x#-intercepts of #y=x^2-4x#?

1 Answer
Apr 18, 2017

# x= 0 and x = 4 #

Explanation:

To find the #x# intercept of the equation #y=x^2-4x#, we input # y=0#, as at the #x# intercept the #y# co-ordinate will be zero.

We get,

#x^2-4x=0#
#x^2=4x#
#x=4#

#x=0# is an obvious answer.

graph{x^2-4x [-3.54, 6.46, -4.22, 0.78]}