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4.3 — More Features, More Functions
In Sections 4.1–4.2 you worked with separate rules defined on different intervals. Today, one important shape—a V—shows how two linear rules can fit together as a single absolute-value function.
1. In class: A Solidify Understanding Task
Absolute value: distance from zero
The absolute value of a number is its distance from zero on the number line. Distance cannot be negative, so \(|4|=4\) and \(|-4|=4\). The bars mean “take the nonnegative magnitude,” not “erase any minus sign anywhere in the expression.”
\(|2-7|=|-5|=5\), while \(-|2-7|=-5\). The negative sign outside the bars changes the sign of the answer.
A useful mathematical rule is that \(|u|=-u\) when \(u<0\), and \(|u|=u\) when \(u\ge0\). This is the foundation of today’s two-piece graphs.
Michelle’s trip: two straight segments, one V
In the packet’s Part I graph, \(t\) is time in minutes and \(D(t)\) is Michelle’s distance from the lake in blocks. She starts six blocks away at \(t=0\), reaches the lake at \((3,0)\), then ends six blocks away at \(t=6\). The slopes are \(-2\) blocks per minute on the way toward the lake and \(+2\) blocks per minute leaving it.
The point-slope form of a line through \((t_1,D_1)\) is \(D-D_1=m(t-t_1)\). Both pieces pass through \((3,0)\):
Before the lake: \(D-0=-2(t-3)\), or \(D=-2(t-3)\), for \(0\le t<3\).
After reaching the lake: \(D-0=2(t-3)\), or \(D=2(t-3)\), for \(3\le t\le6\).
Therefore
\[D(t)=\begin{cases}-2(t-3),&0\le t<3,\\2(t-3),&3\le t\le6.\end{cases}\]Because the left piece reverses the negative difference \(t-3\) and the right piece keeps the positive difference, the same ride can be expressed concisely as \(D(t)=2|t-3|\) for \(0\le t\le6\).
The vertex \((3,0)\) is the turning point. The graph is symmetric about \(t=3\). Equal slopes in magnitude on either side make a symmetric V. Notice that distance from the lake increases on the return; it is not the total distance Michelle has biked.
Building the parent absolute-value graph
Compare the line \(f(x)=x\) with \(g(x)=|x|\). For nonnegative inputs, they are equal. For negative inputs, the absolute-value output has the opposite sign: for example, \(f(-3)=-3\) but \(g(-3)=3\). The part of the line below the x-axis reflects above it; the other part stays in place.
| \(x\) | −4 | −3 | −2 | −1 | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|---|---|---|---|
| \(f(x)=x\) | −4 | −3 | −2 | −1 | 0 | 1 | 2 | 3 | 4 |
| \(g(x)=|x|\) | 4 | 3 | 2 | 1 | 0 | 1 | 2 | 3 | 4 |
Every input has exactly one output. The left and right pieces meet at the included vertex \((0,0)\). Domain: all real numbers. Range: \([0,\infty)\).
Moving a V and rewriting it in two pieces
Horizontal and vertical shifts
Starting with \(y=|x|\), replacing \(x\) by \(x-h\) moves the vertex to the right by \(h\), and adding \(k\) moves it up by \(k\). Thus \(y=|x-1|+2\) has vertex \((1,2)\). To write its pieces, ask where the inside \(x-1\) changes sign: at \(x=1\).
On the left, this simplifies to \(-x+3\); on the right, to \(x+1\). Check the meeting point: either gives \(2\) when \(x=1\). The vertex is \((1,2)\), not \((1,0)\).
Vertical stretches and reflections
Multiplying the absolute-value expression by \(a\) changes the slopes to \(-a\) and \(a\) (in left-to-right order) when \(a>0\). For example, \(3|x+2|+1\) has vertex \((-2,1)\) and rises three units for each horizontal unit away from the vertex. A negative multiplier reflects the V across a horizontal line through its vertex, making an upside-down V.
Given \(f(x)=-3(x+2)+1\) for \(x<-2\), and \(f(x)=3(x+2)+1\) for \(x\ge-2\), the common sign-switch occurs at \(x=-2\). Thus
\[\boxed{f(x)=3|x+2|+1}.\]Its vertex is \((-2,1)\), and its values include \(f(-3)=4\), \(f(-2)=1\), and \(f(-1)=4\).
For \(q(x)=-2|x-4|+3\), the vertex is \((4,3)\). When \(x<4\), \(|x-4|=-(x-4)\), so \(q(x)=2(x-4)+3\). When \(x\ge4\), \(q(x)=-2(x-4)+3\). Both pieces decrease in height as you move away from the vertex; the range is \(( -\infty,3]\).
Putting the ideas together
After studying the individual changes, we can summarize a linear absolute-value function as
\[y=a|x-h|+k.\]Its vertex is \((h,k)\), its left arm has slope \(-a\), and its right arm has slope \(a\). If \(a>0\), it opens upward; if \(a<0\), downward. The piecewise form is
\[y=\begin{cases}-a(x-h)+k,&x<h,\\a(x-h)+k,&x\ge h.\end{cases}\]2. Ready homework (1–6): quadratic x-intercepts
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
An x-intercept is where a graph crosses or touches the x-axis, so its y-value is zero. To find x-intercepts algebraically, set \(y=0\) and solve for \(x\). This is review of quadratic algebra, not a new type of absolute-value function.
Factoring when the equation is expanded
Find the x-intercepts of \(y=x^2+5x-14\). First set \(y=0\):
\[0=x^2+5x-14=(x+7)(x-2).\]By the zero-product property, \(x=-7\) or \(x=2\). The intercepts are \((-7,0)\) and \((2,0)\). Substitution checks both.
Find the zeros of \(y=6x^2+x-2\). Because \(6(-2)=-12\) and \(4+(-3)=1\), split the middle term:
\[6x^2+4x-3x-2=2x(3x+2)-(3x+2)=(2x-1)(3x+2).\]Hence \(x=\frac12\) or \(x=-\frac23\).
When the quadratic is in vertex form
For \(y=2(x-3)^2-8\), set \(y=0\):
\[2(x-3)^2=8\Rightarrow (x-3)^2=4\Rightarrow x-3=\pm2.\]Thus \(x=1\) or \(x=5\). Do not forget both the positive and negative square roots.
For \(y=\frac12(x+2)^2-\frac92\), set \(y=0\). Multiply by 2: \((x+2)^2=9\). Taking square roots yields \(x+2=\pm3\), so \(x=1\) and \(x=-5\).
When there is no real intercept: \((x-2)^2+5=0\) would require a square to equal \(-5\), which is impossible for real \(x\). In that case there are no x-intercepts.
Check your work — Ready 1–6 answer key
- \((-5,0),(2,0)\)
- \((-7,0),(-1,0)\)
- \((-\frac52,0),(\frac43,0)\)
- \((-1,0),(5,0)\)
- \((-6,0),(0,0)\)
- \((-1,0),(3,0)\)
3. Set homework (7–10): tables, graphs, and formulas
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
Read a V from a table (Set 7)
In the packet’s table, the smallest value is \(0\) when \(x=2\), and outputs increase by 3 with each unit away from \(2\). This gives \(f(x)=3|x-2|\), and therefore
\[f(x)=\begin{cases}-3(x-2),&x<2,\\3(x-2),&x\ge2.\end{cases}\]The table supports this simple V-shaped model. As with any finite table, the listed values by themselves do not uniquely determine the function for every real input; the exercise intends an absolute-value model.
Read a formula from a V graph (Set 8)
Locate the vertex \((h,k)\); then count rise over run on one arm. In packet problem 8 the vertex is \((-5,-4)\), and each arm rises one unit per horizontal unit. Therefore the graph represents \(y=|x+5|-4\).
Make a table around the vertex (Set 9)
For \(h(x)=5|x-6|-8\), the vertex is \((6,-8)\). Choose input values on both sides, not just one side:
| \(x\) | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|
| \(5|x-6|-8\) | 2 | −3 | −8 | −3 | 2 |
Graph from a two-piece formula (Set 10)
For \(x<6\), the packet gives \(-\tfrac23(x-6)+4\); for \(x\ge6\), it gives \(\tfrac23(x-6)+4\). Both pieces meet at \((6,4)\). Use the vertex and symmetric points: \((3,6)\), \((6,4)\), \((9,6)\). This is the absolute-value function \(f(x)=\tfrac23|x-6|+4\).
Check your work — Set 7–10 answer key
- \(f(x)=\begin{cases}-3(x-2),&x<2\\3(x-2),&x\ge2\end{cases}\)
- \(f(x)=|x+5|-4\)
- Sample points: \((4,2),(5,-3),(6,-8),(7,-3),(8,2)\); vertex \((6,-8)\).
- Plot/construction not shown. Vertex \((6,4)\); left slope \(-\frac23\), right slope \(\frac23\); passes through \((0,8),(3,6),(9,6),(12,8)\).
4. Go homework (11–20): negatives inside and outside bars
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
The packet’s Go expressions combine unary minus signs and absolute-value bars. The order matters: evaluate the innermost expression first, apply absolute value bars, and then apply any negative sign outside.
| Expression when \(x=-7\) | Work | Value |
|---|---|---|
| \(-x\) | \(-(-7)\) | \(7\) |
| \(|x|\) | \(|-7|\) | \(7\) |
| \(-|x|\) | \(-|-7|\) | \(-7\) |
| \(|-x|\) | \(|-(-7)|=|7|\) | \(7\) |
| \(-|-x|\) | \(-|-(-7)|=-|7|\) | \(-7\) |
Evaluate \(-|x|\) when \(x=12\).
Substitute: \(-|12|\). The absolute value is \(12\), and the leading negative stays outside, so the answer is \(-12\).
Evaluate \(-|-(-x)|\) when \(x=-9\).
Start inside: \(-x=-(-9)=9\). The next negative gives \(-(-x)=-9\). Now apply the bars: \(|-9|=9\). Finally apply the outer negative: \(-9\).
Evaluate \(-|\,-|x-3|+2\,|\) when \(x=-1\).
First \(x-3=-4\), so \(|x-3|=4\). Then \(-4+2=-2\). The outer bars give \(2\). The initial negative makes the result \(-2\).
Check your work — Go 11–20 answer key
- \(-4\)
- \(7\)
- \(0\)
- \(11\)
- \(25\)
- \(25\)
- \(-2\)
- \(6\)
- \(8\)
- \(-9\)
5. Teacher’s review (teacher/parent reference)
Set 7: A finite table is insufficient to establish a unique function globally, though it suggests \(3|x-2|\) and the section makes the intended modeling family clear. A more mathematically precise question would specify that the table is generated by a linear absolute-value function.
Main task: No serious mathematical error was identified in the absolute-value / piecewise representations. Keep the bicycle graph’s specified contextual domain \([0,6]\) distinct from the all-real-number domain of its algebraic extension.
6. Packet coverage — Section 4.3 only
| Packet exercises | Lesson topic |
|---|---|
| Main task Part I, 1–3 | Michelle’s trip, slopes, point-slope forms and a V graph (In class) |
| Part II, 4–8 | \(y=x\) versus \(y=|x|\), piecewise rule, complete input table (In class) |
| Part III, 9–15 | Translations, stretches, graphs, absolute-value and piecewise conversion (In class) |
| Ready, 1–6 | Quadratic x-intercepts by factoring or square roots (Section 2) |
| Set, 7–10 | Tables, graph features, values and equation forms (Section 3) |
| Go, 11–20 | Signs and absolute-value evaluation; three original worked examples (Section 4) |
Curriculum source: Mathematics Vision Project, Secondary Math II, Module 4: More Functions, More Features (2017), Section 4.3, packet pp. 11–16. This independently written lesson covers the assigned section only. The embedded graphs work offline; formatted equations use MathJax and require an internet connection.
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