so, hmmm keeping in mind that, an absolute value expression is in effect a piece-wise function, let's find its "slopes", the negarive slope on the left-side and the positive slope on the right side, and let's use the vertex point as our first point
what is the slope of a line that passes through (2, -1) and (0,7)
[tex]\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ 2}}\quad ,&{{ -1}})\quad
% (c,d)
&({{ 0}}\quad ,&{{ 7}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{7-(-1)}{0-2}\implies \cfrac{7+1}{-2}\implies -4
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-(-1)=-4(x-2)
\\\\\\
\boxed{y+1=-4(x-2)}[/tex]
now, what is the slope of a line that passes through (2, -1) and (4,7)
[tex]\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ 2}}\quad ,&{{ -1}})\quad
% (c,d)
&({{ 4}}\quad ,&{{ 7}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{7-(-1)}{4-2}\implies \cfrac{7+1}{2}\implies 4
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-(-1)=4(x-2)
\\\\\\
\boxed{y+1=4(x-2)}[/tex]
since an absolute value expression is just a piece-wise with a +/- versions, then we can just combine those two.
[tex]\bf \begin{cases}
y+1=-4(x-2)\\
y+1=4(x-2)
\end{cases}\implies y+1=\pm 4(x-2)
\\\\\\
y+1=|4(x-2)|\implies y=|4(x-2)|-1[/tex]
