[tex]\bf (\stackrel{x_1}{6}~,~\stackrel{y_1}{-8})~\hspace{10em} slope = 0 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-(-8)=0(x-6)\implies y+8=0\implies y=-8[/tex]
Answer: Hello there!
we want to find the equation that passes through the point (6, -8) (where this notation stands for (x,y)) with a slope of 0.
a linear equation has the form y = ax+ b
where a is the slope and b is the intercept.
if a = 0, our equation has the form y = b (has no dependence of x)
so we have a constant equation that passes through the point (6, -8) , this means x = 6 and y = -8, but we know that y is constant:
then y = -8 = b
so our equation is y(x) = -8
