How do you find the slope and intercept of #6y+6 = 0#?

1 Answer
Aug 2, 2018

Answer:

The slope is #0#.

There is no #x#-intercept.

The #y#-intercept is at #(0, -1)#.

Explanation:

#6y + 6 = 0#

First, make #y# by itself. Subtract #color(blue)6# from both sides:
#6y + 6 quadcolor(blue)(-quad6) = 0 quadcolor(blue)(-quad6)#

#6y = -6#

Divide both sides by #color(blue)6#:
#(6y)/color(blue)6 = (-6)/color(blue)6#

#y = -1#

When the equation is #y# equals to a number constant, that means it is a horizontal slope, or it has a slope of #0#.

There is no #x#-intercept since the equation never touches the #x#-axis. The #y#-intercept would be at #(0, -1)#, since whatever #x#-value you put into the equation does nothing. The equation is a constant, and it is always where #y = -1#.

Hope this helps!