Question

Write an absolute value equation for the graph shown below

Answer 1

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)

$\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ 2}}\quad ,&{{ -1}})\quad % (c,d) &({{ 0}}\quad ,&{{ 7}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{7-(-1)}{0-2}\implies \cfrac{7+1}{-2}\implies -4 \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-(-1)=-4(x-2) \\\\\\ \boxed{y+1=-4(x-2)}$

now, what is the slope of a line that passes through (2, -1) and (4,7)

$\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ 2}}\quad ,&{{ -1}})\quad % (c,d) &({{ 4}}\quad ,&{{ 7}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{7-(-1)}{4-2}\implies \cfrac{7+1}{2}\implies 4 \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-(-1)=4(x-2) \\\\\\ \boxed{y+1=4(x-2)}$

since an absolute value expression is just a piece-wise with a +/- versions, then we can just combine those two.

$\bf \begin{cases} y+1=-4(x-2)\\ y+1=4(x-2) \end{cases}\implies y+1=\pm 4(x-2) \\\\\\ y+1=|4(x-2)|\implies y=|4(x-2)|-1$