Question 1108345
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The graph of {{{f(x)=abs(x)}}} has its vertex at (0,0); the graph of {{{f(x) = abs(x+3)}}} has its vertex at (-3,0).  The x coordinates of the vertices determine where the function must be separated into pieces.<br>
So we will have one function for x less than -3, another for x greater than or equal to -3 and less than 0, and a third for x greater than or equal to 0.<br>
x < -3: {{{abs(x) = -x}}}; {{{abs(x+3) = -x-3}}}; {{{abs(x)-abs(x+3) = (-x) - (-x-3) = 3}}}<br>
-3 <= x < 0: {{{abs(x) = -x}}}; {{{abs(x+3) = x+3}}}' {{{abs(x)-abs(x+3) = (-x) = (x+3) = -2x-3}}}<br>
x >= 0: {{{abs(x) = x}}}; {{{abs(x+3) = x+3}}}; {{{abs(x)-abs(x+3) = (x) = (x+3) = -3}}}<br>
The piecewise function is
{{{f(x) = 3}}} for x < -3;
{{{f(x) = -2x-3}}} for 3 <= x < 0;
{{{f(x) = -3}}} for x >= 0<br>
The graph:<br>
{{{graph(400,400,-5,5,-5,5,abs(x)-abs(x+3))}}}<br>
The y intercept is when x is 0: (0,-3).<br>
The x intercept(s) are when y is 0.  Either algebraically or from the graph, the only x intercept is (-1.5,0).