How do you find the x and y intercepts for #-4x-7y+7=0#?

2 Answers
Mar 2, 2018

Answer:

I tried this:

Explanation:

You need to set #x=0# and #y=0# in your expression.

Let us find the #x# intercept; we need to, basically, stay on the #x# axis so that #y# has to be zero (otherwise we are not on the #x# axis).
So, we get:

#y=0#
into your expression:
#-4x-7*color(red)(0)+7=0#
#-4x=-7#
#x=7/4#
So the #x-#axis intercept will be:
#(7/4,0)#

Now for the #y# intercept we need #x# to be zero and we get:
#x=0#
into your expression:
#-4*color(red)(0)-7y+7=0#
#-7y=-7#
#y=7/7=1#
So the #y-#axis intercept will be:
#(0,1)#

Let us check and "see" our intercepts graphically:
graph{-(4/7)x+1 [-10, 10, -5, 5]}

Mar 2, 2018

Answer:

#"x-intercept "=7/4," y-intercept "=1#

Explanation:

#"to find the intercepts that is where the graph crosses"#
#"the x an d y axes"#

#• " let x = 0, in the equation for y-intercept"#

#• " let y = 0, in the equation for x-intercept"#

#-4x-7y+7=0#

#rArr-4x-7y=-7#

#x=0rArr0-7y=-7rArry=1larrcolor(red)"y-intercept"#

#y=0rArr-4x+0=-7rArrx=7/4larrcolor(red)"x-intercept"#
graph{(y+4/7x-1)((x-7/4)^2+(y-0)^2-0.04)((x-0)^2+(y-1)^2-0.04)=0 [-10, 10, -5, 5]}