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

Question

Nathan Axel · Accepted Answer

The slope is #0#.

There is no #x#-intercept.

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

#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#. I hope this is useful.

Genesis Allen · Answer

undefined To find the slope-intercept form of the equation $6y + 6 = 0$, first, isolate $y$ by subtracting 6 from both sides, yielding $6y = -6$. Then, divide both sides by 6 to solve for $y$, which results in $y = -1$. The equation is now in slope-intercept form, where the slope is 0 and the y-intercept is -1.