Example 4
Completion requirements
| Example 4 |
Use the graph of \(y = -\frac{1}{2}\left( {x - 3} \right)^2 + 2\) to graph \(y = \left| { - \frac{1}{2}\left( {x - 3} \right)^2 + 2} \right|\).
The function \(y = -\frac{1}{2}\left( {x - 3} \right)^2 + 2\) is written in vertex form, so the vertex occurs at \((3, 2)\). The \(a\)-value is negative, so the parabola opens down. The \(x\)-intercepts can be found by determining the zeros of the function.
The \(x\)-intercepts are \(1\) and \(5\).
The \(y\)-intercept can be determined by setting \(x\) to zero.
Sketch the graph of \(y = -\frac{1}{2}\left( {x - 3} \right)^2 + 2\) using the intercepts, vertex, and direction of opening.
As before, the negative portion of \(y = -\frac{1}{2}\left( {x - 3} \right)^2 + 2\) becomes positive in \(y = \left| { - \frac{1}{2}\left( {x - 3} \right)^2 + 2} \right|\). Either use \(y = -\left(-\frac{1}{2}\left( {x - 3} \right)^2 + 2\right)\) to graph this portion of the function, or pick individual points with negative \(y\)-values, and make those \(y\)-values positive.
\[\begin{align}
0 &= - \frac{1}{2}\left( {x - 3} \right)^2 + 2 \\
-2 &= - \frac{1}{2}\left( {x - 3} \right)^2 \\
4 &= \left( {x - 3} \right)^2 \\
\pm 2 &= x - 3 \\
1,5 &= x \end{align}\]
0 &= - \frac{1}{2}\left( {x - 3} \right)^2 + 2 \\
-2 &= - \frac{1}{2}\left( {x - 3} \right)^2 \\
4 &= \left( {x - 3} \right)^2 \\
\pm 2 &= x - 3 \\
1,5 &= x \end{align}\]
The \(x\)-intercepts are \(1\) and \(5\).
The \(y\)-intercept can be determined by setting \(x\) to zero.
\[\begin{align}
y &= - \frac{1}{2}\left( {x - 3} \right)^2 + 2 \\
y &= - \frac{1}{2}\left( {0 - 3} \right)^2 + 2 \\
y &= - \frac{5}{2} \end{align}\]
y &= - \frac{1}{2}\left( {x - 3} \right)^2 + 2 \\
y &= - \frac{1}{2}\left( {0 - 3} \right)^2 + 2 \\
y &= - \frac{5}{2} \end{align}\]
Sketch the graph of \(y = -\frac{1}{2}\left( {x - 3} \right)^2 + 2\) using the intercepts, vertex, and direction of opening.
As before, the negative portion of \(y = -\frac{1}{2}\left( {x - 3} \right)^2 + 2\) becomes positive in \(y = \left| { - \frac{1}{2}\left( {x - 3} \right)^2 + 2} \right|\). Either use \(y = -\left(-\frac{1}{2}\left( {x - 3} \right)^2 + 2\right)\) to graph this portion of the function, or pick individual points with negative \(y\)-values, and make those \(y\)-values positive.