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What two slopes appear in the graph of \(y = \left|\frac{3}{2}x - 4 \right|\)? For what interval does each slope apply?
The slope of the corresponding function \(y = \frac{3}{2}x - 4\) is \(\frac{3}{2}\). The absolute value function will include this slope and its negative, so the slopes are \(\frac{3}{2}\) and \(-\frac{3}{2}\).
For \(y = \left|\frac{3}{2}x - 4 \right|\), the slope will change at the zero of \(y = \frac{3}{2}x - 4\).\[\begin{align}
y &= \frac{3}{2}x - 4 \\
0 &= \frac{3}{2}x - 4 \\
4 &= \frac{3}{2}x \\
\frac{8}{3} &= x \\
\end{align}\]
The slopes will change sign at \(x = \frac{8}{3}\). The slope of \(y = \frac{3}{2}x - 4\) is positive, so the function increases from left to right on a graph. The left side of the graph of \(y = \frac{3}{2}x - 4\) will be reflected about the \(x\)-axis for the graph of \(y = \left|\frac{3}{2}x - 4 \right|\), so the slope of the left side of the absolute value function becomes \(-\frac{3}{2}\).
The slope of \(y = \left|\frac{3}{2}x - 4 \right|\) will be \(-\frac{3}{2}\) for \(x \lt \frac{8}{3}\), and the slope will be \(\frac{3}{2}\) for \(x \gt \frac{8}{3}\).
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Represent \(f(x) = \left|3 - 2x \right|\) as a piecewise function.
The sub-functions that make up the piecewise function will be \(g(x) = 3 - 2x\) and its negative, \(h(x) = -(3 - 2x)\) or \(h(x) = 2x - 3\).
The intervals in the piecewise function will be separated by the zero of \(f(x)\).\[\begin{align}
f\left( x \right) &= \left| {3 - 2x} \right| \\
0 &= \left| {3 - 2x} \right| \\
0 &= 3 - 2x \\
2x &= 3 \\
x &= \frac{3}{2} \end{align}\]
The intervals will be separated by \(x = \frac{3}{2}\). The slope of \(g(x) = 3 - 2x\) is negative, so the function decreases from left to right on a graph. As such, for \(x\)-values greater than \(\frac{3}{2}\), \(g(x)\) will be negative, and \(h(x)\) and \(f(x)\) will be positive.\[f\left( x \right) = \left\{ \begin{align}
3 - 2x{\rm \thinspace { for }} \thinspace x \le \frac{3}{2} \\
2x - 3{\rm \thinspace { for }} \thinspace x > \frac{3}{2} \end{align} \right.\]
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Consider the function \(y = \left|x^2 + x - 6 \right|\).
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Determine the zeros of \(y = \left|x^2 + x - 6 \right|\).
The \(x\)-intercepts are invariant, so the zeros are the same for \(y = \left|x^2 + x - 6 \right|\) and \(y = x^2 + x -6\).
\(\begin{align}
y &= x^2 + x - 6 \\
0 &= x^2 + x - 6 \\
0 &= \left( {x + 3} \right)\left( {x - 2} \right) \end{align}\)\(\begin{align}
0 &= x + 3 \\
-3 &= x \end{align}\)\(\begin{align}
0 &= x - 2 \\
2 &= x \end{align}\)
The zeros are \((-3, 0)\) and \((2, 0)\).
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Determine the \(y\)-intercept of \(y = \left|x^2 + x - 6 \right|\).
\(\begin{align}
y &= \left| {x^2 + x - 6} \right| \\
y &= \left| {0^2 + 0 - 6} \right| \\
y &= \left| { - 6} \right| \\
y &= 6 \\
\end{align}\)
The \(y\)-intercept occurs at \((0, 6)\).
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Sketch the graph of \(y = \left|x^2 + x - 6 \right|\).
Use the graph of \(y = x^2 + x - 6\) to graph \(y = \left|x^2 + x - 6 \right|\).
- State the domain and range of \(y = \left| x^2 + x - 6 \right|\).
Domain: {\(x | x \in \rm{R}\)}
Range: {\(y | y \ge 0, \thinspace y \in \rm{R}\)}
- Explain
how the graph of the function shown can be used to represent the
distance a man is from the \(6^{th}\) floor, as he rides an elevator
from the ground floor to the \(17^{th}\) floor of a building.
At the ground floor, the man is five floors away from the sixth floor. As the elevator rises, the distance to the sixth floor decreases, which is represented by the left portion of the graph. When he passes the sixth floor, the distance to the sixth floor is zero, which is represented by the \(x\)-intercept. As the elevator continues to rise, the man gets farther away from the sixth floor, which is represented by the right portion of the graph.
| For further information about absolute value and absolute value functions, see pp. 358 – 363 and pp. 368 – 375 of Pre-Calculus 11. |