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  1. Solve \(\frac{1}{2}(x - 3)^2 + 2 = \left| 2x - 6\right|\) graphically.  Verify any solutions by substitution.

    Graph the functions \(f(x) = \frac{1}{2} (x - 3)^2 + 2\) and \(g(x) = \left|2x - 6\right|\).


    The two curves intersect at \((1, 4)\) and \((5, 4)\), so the solutions to the equation are \(1\) and \(5\).

    Verify for \(x = 1\).

    Left Side Right Side
    \[\begin{array}{r}
     \frac{1}{2}\left( {x - 3} \right)^2 + 2 \\
     \frac{1}{2}\left( {\left( 1 \right) - 3} \right)^2 + 2 \\
     4  \end{array}\]    		 \[\begin{array}{l}
     \left| {2x - 6} \right| \\
     \left| {2\left( 1 \right) - 6} \right| \\
     \left| { - 4} \right| \\
     4  \end{array}\]
    \(\hspace{30pt}\)LS = RS

    Verify for \(x = 5\).

    Left Side Right Side
    \[\begin{array}{r}
     \frac{1}{2}\left( {x - 3} \right)^2 + 2 \\
     \frac{1}{2}\left( {\left( 5 \right) - 3} \right)^2 + 2 \\  4 \end{array}\]    		 \[\begin{array}{l}
     \left| {2x - 6} \right| \\
     \left| {2\left( 5 \right) - 6} \right| \\
     \left| 4 \right| \\
     4 \end{array}\]
    \(\hspace{30pt}\)LS = RS


  2. Solve \(\left| 1 - 3x\right| = x - 1\) graphically.

    Graph the functions \(f(x) = \left|1 - 3x\right|\) and \(g(x) = x = -1\).


    There are no points of intersection, so there is no solution to the equation.