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4.4 — Reflections of a Bike Lover
In Section 4.3 you expressed a V-shaped absolute-value graph with two linear pieces. Today the same idea works with curved functions too. We first learn to change the output values on a graph, and then write a piecewise equation that captures the change.
1. In class: A Practice Understanding Task
What does absolute value do to an output?
For any real number \(u\), its absolute value \(|u|\) is its nonnegative distance from zero. If a function \(f(x)\) has a positive or zero output, \(|f(x)|=f(x)\). If it has a negative output, \(|f(x)|=-f(x)\). So the new function \(g(x)=|f(x)|\) keeps points on or above the x-axis and reflects points below the x-axis upward.
Do not confuse \(|f(x)|\) with \(f(|x|)\): the first takes absolute value of the output; the second changes the input.
Suppose the original graph passes through \((-3,5)\), \((0,-4)\), and \((2,0)\). On the graph of \(y=|f(x)|\), these become \((-3,5)\), \((0,4)\), and \((2,0)\). Only the point below the x-axis moves.
Reflecting a parabola: main task 1–4
Start with \(f(x)=x^2-4\), an upward-opening parabola with vertex \((0,-4)\). To find its x-intercepts set \(x^2-4=0\), so \(x^2=4\), giving \(x=-2\) or \(x=2\). Between the intercepts, the parabola lies below the x-axis; outside them, it lies on or above it.
Because \(x^2-4<0\) precisely when \(-2<x<2\), we negate the quadratic only on that interval:
\[g(x)=|x^2-4|=\begin{cases}x^2-4,&x\le-2,\\4-x^2,&-2<x<2,\\x^2-4,&x\ge2.\end{cases}\]Check: \(g(0)=|0-4|=4\), agreeing with the reflected center. At both boundary points the output is zero, so each can be assigned to either neighboring piece (but avoid assigning different answers to the same input). The new graph has local minima at \((-2,0)\) and \((2,0)\), and a local maximum at \((0,4)\).
A reliable procedure for any \(|f(x)|\)
Step 1. Identify the underlying function \(f(x)\), ignoring the outer bars temporarily. Step 2. Find where \(f(x)=0\); those points separate possible sign regions. Step 3. Determine whether \(f(x)\) is positive or negative in each interval, using a simple test point or graph. Step 4. Write \(f(x)\) on nonnegative intervals and \(-f(x)\) on negative intervals. Step 5. Check the boundary values and domain.
Take \(f(x)=(x+1)^2-9\). Set it equal to zero:
\[(x+1)^2=9\Rightarrow x+1=\pm3\Rightarrow x=-4\ \text{or}\ x=2.\]Test \(x=-5\): \(f(-5)=16-9=7>0\). Test \(x=0\): \(f(0)=1-9=-8<0\). Test \(x=3\): \(f(3)=16-9=7>0\). Therefore the sign pattern is positive, negative, positive. The turning point of the original parabola is \((-1,-9)\), which reflects to \((-1,9)\).
Shifted and raised reflection examples
Main task 5–8: a shifted parabola
For \(g(x)=|(x+1)^2-9|\), the x-intercepts are \(-4\) and \(2\). The center segment reflects upward, and the outer segments remain unchanged.
Main task 10: vertical shift outside the bars
In \(g(x)=|x^2-4|+1\), first reflect the negative center of \(x^2-4\), then shift every output up 1. The breakpoints stay \(-2\) and \(2\), but the zeros of \(|x^2-4|+1\) disappear: the minimum output is 1.
The outer formula simplifies from \((x^2-4)+1\); the middle formula from \(-(x^2-4)+1\). The key plotted points are \((-2,1)\), \((0,5)\), and \((2,1)\).
Main task 11: two horizontal breakpoints
For \(g(x)=|(x+2)^2-4|+3\), solve \((x+2)^2-4=0\): \(x=-4\) and \(x=0\). Inside \((-4,0)\), negate the quadratic; outside, keep it. Then add 3 everywhere.
Important graph features are \((-4,3)\), \((-2,7)\), \((0,3)\). The range is \([3,\infty)\).
An exponential function can be reflected too (main task 12)
For \(g(x)=|2^x-4|\), first study \(f(x)=2^x-4\). This curve increases as \(x\) increases, and it crosses the x-axis where \(2^x=4=2^2\), so \(x=2\). For \(x<2\), \(2^x<4\), so the original outputs are negative. For \(x\ge2\), they are nonnegative.
Check \(g(0)=|1-4|=3\), \(g(2)=0\), and \(g(3)=4\). Because \(2^x\) approaches 0 but never reaches it as \(x\) goes left, the left arm approaches \(y=4\) without reaching it on that interval. The entire function has domain \(( -\infty,\infty)\) and range \([0,\infty)\).
2. Ready (1–3): reflect geometric figures across \(y=x\)
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.
Unlike reflecting a graph above the x-axis, reflecting a geometric figure across the diagonal line \(y=x\) interchanges the coordinates. This is a separate review skill: \((x,y)\mapsto(y,x)\).
The packet gives \(A(-4,-2)\), \(B(-6,3)\), and \(C(-3,5)\). Swapping coordinates produces \(A'(-2,-4)\), \(B'(3,-6)\), \(C'(5,-3)\). Connect the image points in the same order.
Each connecting segment \(AA'\), \(BB'\), and \(CC'\) is perpendicular to \(y=x\), and \(y=x\) bisects each segment. They are parallel to each other (unless a point already lies on the mirror line, in which case its segment has zero length).
Choose a pentagon in quadrant IV with vertices \(P(1,-2)\), \(Q(4,-1)\), \(R(6,-3)\), \(S(4,-6)\), \(T(1,-5)\). Reflect each by swapping coordinates:
\(P'(-2,1)\), \(Q'(-1,4)\), \(R'(-3,6)\), \(S'(-6,4)\), \(T'(-5,1)\). The image is in quadrant II.
The packet lists \(A(-6,2)\), \(B(-4,5)\), \(C(-2,3)\), \(D(-3,-1)\). Its reflections are \(A'(2,-6)\), \(B'(5,-4)\), \(C'(3,-2)\), \(D'(-1,-3)\).
Check your work — Ready 1–3 answer key
- \(A'=(-2,-4),B'=(3,-6),C'=(5,-3)\). Plot not shown. The connecting segments are parallel to one another and perpendicular to \(y=x\); the reflection line bisects each segment.
- One example: \(A(1,-1),B(4,-1),C(5,-3),D(3,-5),E(1,-3)\) reflect to \(A'(-1,1),B'(-1,4),C'(-3,5),D'(-5,3),E'(-3,1)\). Plot not shown.
- \(A'(2,-6),B'(5,-4),C'(3,-2),D'(-1,-3)\)
3. Set (4–7): sound waves, tables and restricted domains
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.
Sound-wave graph (Set 4)
The packet shows a wave that crosses the x-axis repeatedly. To graph \(y=|f(x)|\), leave the crests above the axis in place; reflect each trough upward. The result has no negative y-values and touches the axis wherever the original wave crossed it. The packet uses “amplitude” for the distance from the undisturbed level; the displayed graph is the time-varying absolute displacement, rather than the single maximum-amplitude number sometimes used in physics.
The quadratic table (Set 5–6)
The packet uses \(g(x)=(x+3)^2-9\) with \(h(x)=|g(x)|\). For every table entry, keep x unchanged and replace a negative g-value by its positive opposite.
| \(x\) | −8 | −7 | −6 | −4 | −3 | −2 | −1 | 0 | 1 | 2 |
|---|---|---|---|---|---|---|---|---|---|---|
| \(g(x)\) | 16 | 7 | 0 | −5 | −8 | −9 | −8 | 0 | 7 | 16 |
| \(h(x)\) | 16 | 7 | 0 | 5 | 8 | 9 | 8 | 0 | 7 | 16 |
Graph these points and join them with the appropriate curved pieces. The zeros remain at \(x=-6\) and \(x=0\); the middle trough of \(g\), at \((-3,-9)\), becomes a high point \((-3,9)\) on \(h\).
Restricted-domain piecewise form (Set 7)
The requested domain is \([-8,2]\), not every real x. Solve \((x+3)^2-9=0\): \(x=-6\) or \(x=0\). Thus:
Domain: \([-8,2]\). Range: \([0,16]\). The largest y-value on this domain is 16 at the endpoints \(x=-8\) and \(x=2\).
Check your work — Set 4–7 answer key
- Plot/construction not shown. Keep portions above the time axis unchanged, reflect portions below it upward, and keep all zeros fixed; the resulting outputs are nonnegative.
- Using the printed table, change the outputs at \(x=-4,-3,-2,-1\) from \(-5,-8,-9,-8\) to \(5,8,9,8\). Packet error: this table does not match \(g(x)=(x+3)^2-9\) (for example, \(g(-3)=-9\), not \(-8\)). For that equation, negative outputs occur at \(x=-5,-4,-3,-2,-1\) and become \(5,8,9,8,5\).
- Plot/construction not shown. Using the stated equation, \(h(x)=|(x+3)^2-9|\) has zeros \((-6,0),(0,0)\) and a middle maximum at \((-3,9)\); reflect the original portion between the zeros upward. The printed table is inconsistent with this equation.
- \(h(x)=\begin{cases}(x+3)^2-9,&-8\le x\le-6\\9-(x+3)^2,&-6<x<0\\(x+3)^2-9,&0\le x\le2\end{cases}\). Uses the stated equation rather than the erroneous table.
4. Go (8–14): simplifying radicals
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.
These are independent algebra review problems. A radical is in simplest radical form when it contains no perfect-square factors larger than 1. For example, \(\sqrt{72}=\sqrt{36\cdot2}=6\sqrt2\). Add or subtract radicals only when the expressions under their roots are the same.
Combine matching radical terms
Simplify \((5-2\sqrt7)-(8+3\sqrt7)\).
\[5-2\sqrt7-8-3\sqrt7=\boxed{-3-5\sqrt7}.\]The minus in front of parentheses changes both terms inside.
Extract perfect squares before combining
Simplify \(\sqrt{75}-2\sqrt{12}+\sqrt{27}\).
\[=5\sqrt3-2(2\sqrt3)+3\sqrt3=\boxed{4\sqrt3}.\]All three terms become like radical terms after simplification.
Simplify \((2-\sqrt{50})+(6+2\sqrt8)-(3-\sqrt{18})\).
First \(\sqrt{50}=5\sqrt2\), \(\sqrt8=2\sqrt2\), and \(\sqrt{18}=3\sqrt2\). Distribute the signs:
\[2-5\sqrt2+6+4\sqrt2-3+3\sqrt2=\boxed{5+2\sqrt2}.\]Radicals with variables and absolute values
For real \(b\), \(\sqrt{b^2}=|b|\), not always \(b\). Thus \(\sqrt{3b^2}=|b|\sqrt3\) and \(\sqrt{27b^2}=3|b|\sqrt3\). When no sign restriction on \(b\) is supplied, the absolute-value bars matter.
Simplify \((3r^2-8\sqrt{3b^2})-(2r^2-3\sqrt{27b^2})\).
\[=r^2-8|b|\sqrt3+9|b|\sqrt3=\boxed{r^2+|b|\sqrt3}.\]Change radical notation into exponent notation
The square root is the one-half power: \(\sqrt{x^n}=x^{n/2}\) when \(x\ge0\). More generally, for odd positive n, \(\sqrt{x^n}=x^{(n-1)/2}\sqrt x\) when \(x\ge0\). For example, \(\sqrt{x^7}=x^3\sqrt x=x^{7/2}\).
Write \(\sqrt x+\sqrt{x^3}+\sqrt{x^5}\) using exponents, assuming \(x\ge0\):
\[\boxed{x^{1/2}+x^{3/2}+x^{5/2}}.\]The actual packet continues the same pattern through \(\sqrt{x^{15}}\).
Extra original Go practice (fully worked)
Simplify \(\sqrt{48}+\sqrt{27}\).
\[\sqrt{16\cdot3}+\sqrt{9\cdot3}=4\sqrt3+3\sqrt3=\boxed{7\sqrt3}.\]Simplify \((4-2\sqrt{75})-(7+\sqrt{12})+3\sqrt{27}\).
Rewrite \(\sqrt{75}=5\sqrt3\), \(\sqrt{12}=2\sqrt3\), \(\sqrt{27}=3\sqrt3\), then distribute:
\[4-10\sqrt3-7-2\sqrt3+9\sqrt3=\boxed{-3-3\sqrt3}.\]For real \(t\), simplify \(2\sqrt{12t^2}-\sqrt{27t^2}\); then, assuming \(t\ge0\), rewrite \(\sqrt{t^{11}}\) as a power.
\[2(2|t|\sqrt3)-3|t|\sqrt3=\boxed{|t|\sqrt3},\qquad \sqrt{t^{11}}=\boxed{t^{11/2}}.\]Check your work — Go 8–14 answer key
- \(-1+6\sqrt5\)
- \(1-6\sqrt{13}\)
- \(-1+16\sqrt2\)
- \(9\sqrt2\)
- \(-77+3\sqrt5\)
- \(r^2+|b|\sqrt3\) (because \(\sqrt{b^2}=|b|\)).
- \(x^{\frac12}+x^{\frac32}+x^{\frac52}+x^{\frac72}+x^{\frac92}+x^{\frac{11}2}+x^{\frac{13}2}+x^{\frac{15}2}\)
5. Teacher’s review
Go 13 — missing sign assumption. Simplifying the printed terms containing \(\sqrt{3b^2}\) and \(\sqrt{27b^2}\) requires absolute values unless \(b\ge0\) is explicitly assumed. Recommend adding “Assume \(b\ge0\)” or accepting the mathematically general answer with \(|b|\).
Set 4 — terminology. Calling every instantaneous distance from the axis the “amplitude” may conflict with the conventional scientific meaning of maximum wave displacement. Recommend asking students to graph the absolute value of the wave’s displacement.
6. Section 4.4 coverage
| Packet problems | Explicit instruction |
|---|---|
| Main task 1–4 | \(|x^2-4|\), reflection and three-piece equation (In class) |
| Main task 5–9 | Shifted quadratic, sign intervals, three-piece equation (In class) |
| Main task 10–11 | Outside vertical shifts, shifted quadratic pieces (In class) |
| Main task 12 | Exponential absolute value \(|2^x-4|\) (In class) |
| Ready 1–3 | Coordinate reflections across \(y=x\) (Section 2) |
| Set 4–7 | Sound-wave reflection, table adjustments, quadratic graph and restricted domain (Section 3) |
| Go 8–14 | Like radicals, distributing negatives, square factors, variable radicals and fractional exponents; three original worked problems (Section 4) |
Curriculum source: Mathematics Vision Project, Secondary Math II, Module 4: More Functions, More Features (2017), Section 4.4, packet pp. 17–21. This independently created lesson covers Section 4.4 only. Graphs are embedded; MathJax mathematical typesetting requires an internet connection.
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