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4.7 — More Features, More Functions
The last lesson of the module brings earlier skills together. Rather than learning a new master formula, you will practice making precise statements about functions and choosing equations to meet given conditions.
1. In class: A Practice Understanding Task
How to describe a function from a graph
A function matches each allowed input with exactly one output. Read the graph from left to right. The domain is the set of horizontal coordinates that occur; the range is the set of vertical coordinates. An x-intercept occurs where the output is zero; a y-intercept occurs where the input is zero.
A maximum is the greatest output of a function on its domain, if one exists, and a minimum is the least. Report the location (input) as well as the value (output). A function is increasing when higher input values give higher outputs and decreasing when higher input values give lower outputs. State each interval in terms of x-values, not y-values.
| Feature | What to ask |
|---|---|
| Equation | Which family fits the shape? What shifts, stretches, or slopes are visible? |
| Domain | How far left and right does the actual curve extend? |
| Range | What vertical outputs are reached? |
| Intercepts | Where does the graph meet the horizontal and vertical axes? |
| Extrema | Are there highest or lowest actual values, or merely a boundary of the viewing window? |
| Increasing/decreasing | On what open intervals does the curve rise or fall moving right? |
| Inverse | Exchange every point \((a,b)\) with \((b,a)\); reflect across \(y=x\). |
For \(q(x)=-(x-2)^2+4\), the vertex is \((2,4)\). Because the coefficient of the squared term is negative, the graph opens down. The domain is all real numbers and the range is \(( -\infty,4]\). The maximum location is \(x=2\), its value is 4, and no minimum exists. It increases on \(( -\infty,2)\) and decreases on \((2,\infty)\).
To find roots: \(0=-(x-2)^2+4\), so \((x-2)^2=4\), giving \(x=0\) and \(x=4\). Its x-intercepts are \((0,0)\) and \((4,0)\); its y-intercept is also \((0,0)\).
Recognize common graph families (main task Part I, 1–3)
Square-root curve
A graph with a left endpoint, then a rising curve that gradually flattens, often comes from a square root. For \(y=\sqrt{x+2}-2\), the inside quantity must be nonnegative: \(x+2\ge0\), so \(x\ge-2\). The starting point is \((-2,-2)\); the range is \([-2,\infty)\). It crosses the x-axis when \(\sqrt{x+2}=2\), or \(x=2\). At \(x=0\), the y-intercept is \(\sqrt2-2\). This function rises throughout its domain and has a minimum \(-2\) at \(x=-2\). It has no maximum.
Start with \(y=\sqrt{x+2}-2\). Add 2, then square: \((y+2)^2=x+2\). Thus \(x=(y+2)^2-2\) and \(f^{-1}(x)=(x+2)^2-2\) with the inverse domain \(x\ge-2\). The inverse has range \([-2,\infty)\). Keep the restriction: without it, the whole upward parabola would include an extra branch not present in the inverse.
Exponential decay
For \(f(x)=4(\frac12)^x\), the y-intercept is 4, the graph passes through \((1,2)\), and every step right multiplies the output by \(\frac12\). Its domain is all real numbers, its range is \((0,\infty)\), and it decreases everywhere. It has neither a maximum nor a minimum; the line \(y=0\) is a horizontal asymptote, not an x-intercept.
To reverse this relationship visually, swap points such as \((0,4)\) and \((1,2)\), obtaining inverse points \((4,0)\) and \((2,1)\). An inverse equation requires logarithms, which are beyond today's algebraic requirements; the graph and swapped coordinates suffice.
Absolute-value V shape
The graph \(y=|x+2|-4\) has vertex \((-2,-4)\), domain all reals, range \([-4,\infty)\), and x-intercepts \((-6,0)\) and \((2,0)\). Its y-intercept is \((0,-2)\). It decreases for \(x<-2\), then increases for \(x>-2\). It has a minimum of \(-4\) and no maximum.
Reflect each point of the V shape across \(y=x\). The original vertex \((-2,-4)\) becomes \((-4,-2)\), while the original roots \((-6,0)\) and \((2,0)\) become \((0,-6)\) and \((0,2)\). Since the resulting relation has two outputs at input 0, the unrestricted inverse is not a function. You can graph the entire inverse relation; restrict the original to one arm if asked for an inverse function.
Each graph’s axis has a one-unit scale, but the plotted curves do not consistently show endpoint dots or arrowheads. Unless your teacher specifies that the curves are restricted to the portion inside the viewing box, use the evident parent-function shapes and their natural domains for the smooth and V-shaped graphs. Do not assume a screen edge is a maximum, minimum, or included endpoint. For a piecewise graph, check whether endpoints are explicitly defined before claiming its exact domain.
Describe a piecewise graph (main task Part I, 4)
For a straight segment between two marked points, use \(m=(y_2-y_1)/(x_2-x_1)\), followed by \(y-y_1=m(x-x_1)\). Make a separate rule for each segment and tell which x-values use that rule. A horizontal segment has slope zero. A change from increasing to decreasing creates a local maximum; but whether it is the overall maximum depends on the entire domain.
In the illustrated piecewise graph, key joins are \((-2,-1)\), \((2,5)\), and \((4,1)\). A model consistent with the drawn straight pieces is
Compute the slope of each segment:
From \((-8,-4)\) to \((-2,-1)\), \(m=3/6=\frac12\), hence \(y=\frac12x\). From \((-2,-1)\) to \((2,5)\), \(m=6/4=\frac32\), hence \(y=\frac32x+2\). From \((2,5)\) to \((4,1)\), \(m=-4/2=-2\), hence \(y=-2x+9\). After x=4, the last piece stays at \(y=1\).
If these rules are continued outside the plotted view in the same way, one possible full-domain definition is:
\[f(x)=\begin{cases}\frac12x,&x<-2\\\frac32x+2,&-2\le x<2\\-2x+9,&2\le x<4\\1,&x\ge4.\end{cases}\]The graph rises on \(( -\infty,2)\), falls on \((2,4)\), then is constant on \((4,\infty)\). The local maximum is \(5\) at \(x=2\); with the indicated first branch continued indefinitely to the left, the full function has no global minimum.
To sketch its inverse, flip the coordinates of all junctions: \((-2,-1)\to(-1,-2)\), \((2,5)\to(5,2)\), and \((4,1)\to(1,4)\). The horizontal piece \(y=1\) becomes a vertical ray \(x=1\), so the inverse relation is not a function.
Build a function with specified features (main task Part II, 5–13)
Many sets of features do not determine one unique function. The packet asks for two different examples for each property. First choose a familiar family, then check the desired domain, range, vertex, intercepts, symmetry, and direction.
| Requirement | Example 1 | Example 2 | Why both work |
|---|---|---|---|
| 5. Always increasing | \(f(x)=x\) | \(g(x)=x^3\) | Outputs rise as x rises. |
| 6. Symmetry about y-axis | \(x^2\) | \(|x|\) | Both satisfy \(f(-x)=f(x)\). |
| 7. Minimum −2 at x=5 | \((x-5)^2-2\) | \(|x-5|-2\) | Both have vertex \((5,-2)\). |
| 8. Falls before x=−3; rises after | \((x+3)^2\) | \(|x+3|\) | Both have a turning point at −3. |
| 9. No real roots | \(x^2+1\) | \(-x^2-2\) | Neither graph crosses the x-axis. |
| 10. Domain \([3,\infty)\), range \([0,\infty)\) | \(\sqrt{x-3}\) | \((x-3)^2\), \(x\ge3\) | Both start at \((3,0)\) and produce all nonnegative values. |
| 11. Constant rate of change | \(2x+1\) | \(-3x+4\) | Slope is fixed. |
| 12. Constant second differences | \(x^2\) | \(2x^2+1\) | Equally spaced inputs give constant second differences. |
| 13. Inverse is a function | \(x+1\) | \(x^3\) | Both are one-to-one. |
Find two functions with minimum value 3 at x=−1. Start with a nonnegative expression that becomes zero at \(x=-1\): either \((x+1)^2\) or \(|x+1|\). Adding 3 moves the minimum to y=3. Thus \(f(x)=(x+1)^2+3\) and \(g(x)=|x+1|+3\). The first is a parabola and the second is a V shape. Both have domain all reals and range \([3,\infty)\).
2. Ready — geometric symbols (1–7)
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.
Geometry notation tells you what type of object to draw. Use straightedge-style sketches and label every relevant point.
| Notation in packet | What your sketch must show |
|---|---|
| 1. \(\triangle RST\) | A triangle with vertices R, S, and T. |
| 2. \(\overrightarrow{AB}\) | A ray with endpoint A, passing through B, continuing beyond B. |
| 3. \(\angle XYZ\) | Two rays from vertex Y, one through X and one through Z. |
| 4. \(\overleftrightarrow{GH}\) | A full line passing through G and H, arrows both ways. |
| 5. \(\overline{JK}\perp\overline{PQ}\) | Two segments meeting at a right angle; label the 90° angle. |
| 6. S bisects \(\overline{MN}\) | Points M, S, N lie in order on one segment and \(MS=SN\). |
| 7. \(\overrightarrow{AB}\) bisects \(\angle XYZ\) | Ray AB divides angle XYZ into equal angles, so A must coincide with vertex Y for a literal angle-bisector ray, or the naming requires clarification. |
A ray that bisects an angle must start at the angle’s vertex. The packet names ray \(\overrightarrow{AB}\) and angle \(XYZ\), whose vertex is Y, without specifying how A is related to Y. To sketch the situation as written, place A at Y (same location), let B lie inside angle XYZ, and mark \(\angle XYB\cong\angle BYZ\). A cleaner instruction would name \(\overrightarrow{YB}\) as the bisector.
Check your work — Ready 1–7 answer key
- Plot/construction not shown. Triangle with vertices R, S, T and sides \(\overline{RS},\overline{ST},\overline{TR}\).
- Plot/construction not shown. Ray \(\overrightarrow{AB}\) starts at A, passes through B, and continues beyond B.
- Plot/construction not shown. Angle with vertex Y and sides \(\overrightarrow{YX},\overrightarrow{YZ}\).
- Plot/construction not shown. Line through G and H with arrows at both ends.
- Plot/construction not shown. Segments JK and PQ intersect at a right angle; mark \(90^\circ\).
- Plot/construction not shown. S lies on segment MN and \(MS=SN\); mark the two halves congruent.
- Plot/construction not shown. Ray AB must start at Y (so A coincides with Y), lie inside \(\angle XYZ\), and divide it into equal angles \(\angle XYB=\angle BYZ\). The packet uses A for the ray’s endpoint but Y for the angle’s vertex; those labels must refer to the same point.
3. Set — analyze and construct functions (8–20)
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.
Set 8–10: features from graphs and a piecewise rule
Problem 8: The pictured downward-opening parabola has its vertex at approximately \((0,-1)\) and follows \(y=-x^2-1\). It has domain all reals; range \(( -\infty,-1]\); no x-intercepts; y-intercept \((0,-1)\); maximum \(-1\) at x=0; increases on \(( -\infty,0)\) and decreases on \((0,\infty)\).
Problem 9: The upward V has vertex \((1,3)\), corresponding to \(y=|x-1|+3\). Domain all reals, range \([3,\infty)\), no x-intercepts, y-intercept \((0,4)\), minimum 3 at x=1; decreasing for x<1 and increasing for x>1.
The packet defines \(f(x)=-(x+3)\) for \(x<-3\) and \(f(x)=x+3\) for \(x\ge-3\). These two pieces combine into \(f(x)=|x+3|\). Its domain is \(( -\infty,\infty)\), range \([0,\infty)\); its only x-intercept is \((-3,0)\), its y-intercept is \((0,3)\); its minimum value is 0 at x=−3; it decreases on \(( -\infty,-3)\) and increases on \((-3,\infty)\), with no maximum.
Set 11–20: supply an equation
| Packet prompt | One valid function and the key check |
|---|---|
| 11. Always decreasing | \(f(x)=-2x\), a negative constant slope. |
| 12. Symmetric about \(x=3\) | \(f(x)=(x-3)^2\). |
| 13. Minimum 5 at x=1 | \(f(x)=(x-1)^2+5\). |
| 14. Increasing before 2, decreasing after 2 | \(f(x)=-(x-2)^2\). |
| 15. Exactly one real root | \(f(x)=x-1\). |
| 16. Domain \([-2,\infty)\) | \(f(x)=\sqrt{x+2}\). |
| 17. Range \([0,\infty)\) | \(f(x)=x^2\). |
| 18. Common factor of 2 | \(f(x)=2x^2+4x\); both terms share factor 2. |
| 19. Also a geometric sequence | \(a(n)=3\cdot2^n\), with integer \(n\ge0\). |
| 20. x-intercepts −1 and 1 | \(f(x)=(x+1)(x-1)=x^2-1\). |
For Set 19, a geometric sequence is a function with a discrete domain of integers and a fixed ratio between consecutive terms. Other exponential-type expressions are possible.
Check your work — Set 8–20 answer key
- a. Domain \(\mathbb R\), range \((-\infty,-1]\). b. No x-intercepts; y-intercept \((0,-1)\). c. Maximum \(-1\) at \(x=0\); no minimum. d. Increasing on \((-\infty,0)\), decreasing on \((0,\infty)\).
- a. Domain \(\mathbb R\), range \([3,\infty)\). b. No x-intercepts; y-intercept \((0,4)\). c. Minimum \(3\) at \(x=1\); no maximum. d. Decreasing on \((-\infty,1)\), increasing on \((1,\infty)\).
- a. Domain \(\mathbb R\), range \([0,\infty)\). b. x-intercept \((-3,0)\), y-intercept \((0,3)\). c. Minimum \(0\) at \(x=-3\); no maximum. d. Decreasing on \((-\infty,-3)\), increasing on \((-3,\infty)\).
- One example: \(f(x)=-x\).
- One example: \(f(x)=(x-3)^2\).
- One example: \(f(x)=(x-1)^2+5\).
- One example: \(f(x)=-(x-2)^2\).
- One example: \(f(x)=x^2\).
- One example: \(f(x)=\sqrt{x+2}\).
- One example: \(f(x)=x^2\).
- One example: \(f(x)=2x+2=2(x+1)\).
- One example: \(f(n)=2^{n-1},\ n=1,2,3,\ldots\); common ratio 2.
- One example: \(f(x)=x^2-1\).
4. Go — inverse functions and absolute-value pieces (21–29)
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.
Inverses and domain restriction (21–25)
To find an inverse, write \(y=f(x)\), exchange the variables’ roles, and solve for the new output. For quadratics, the inverse relation normally has two branches, so limit the original domain to the left or right of the vertex if the inverse must be a function.
Find the inverse of \(p(x)=5x-7\).
\[y=5x-7\quad\Rightarrow\quad y+7=5x\quad\Rightarrow\quad x=\frac{y+7}{5}.\]Therefore \(p^{-1}(x)=\frac{x+7}{5}\). Check: \(p(3)=8\) and \(p^{-1}(8)=3\).
Let \(q(x)=(x-4)^2+2\). Find an inverse function.
The unrestricted inverse relation is \(x=4\pm\sqrt{y-2}\), so two outputs would be possible. Choose the original domain \(x\ge4\), giving \(q^{-1}(x)=4+\sqrt{x-2}\). The inverse domain is \([2,\infty)\), range is \([4,\infty)\). Alternatively, choosing original x≤4 gives \(4-\sqrt{x-2}\).
Given \(F=\{(-1,2),(0,4),(3,7)\}\), find its inverse.
Swap coordinates in every pair: \(F^{-1}=\{(2,-1),(4,0),(7,3)\}\). Since each inverse input occurs only once, this inverse relation is also a function.
| Packet item | Inverse result or method |
|---|---|
| 21. \(f(x)=x^2\) | Restrict original x≥0: \(f^{-1}(x)=\sqrt x\) (or x≤0 for \(-\sqrt x\)). |
| 22. \(g(x)=2x+4\) | \(g^{-1}(x)=(x-4)/2\). |
| 23. \(f(x)=(x+1)^2\) | Restrict original x≥−1: \(f^{-1}(x)=\sqrt x-1\); the other branch uses x≤−1. |
| 24. \(h(x)=x/3+6\) | \(h^{-1}(x)=3(x-6)=3x-18\). |
| 25. Ordered pairs | Interchange each x and y; check distinct original outputs to decide if inverse is a function. |
For item 25, the packet’s ordered pairs have distinct output coordinates 5, −9, −2, −5, −4, 6, 10, 8; therefore the inverse is a function. Its inverse pairs, in printed order, are \((5,-3),(-9,-2),(-2,-1),(-5,0),(-4,1),(6,2),(10,3),(8,4)\).
Turn absolute values into piecewise rules (26–29)
The definition \(|A|=A\) for \(A\ge0\), and \(|A|=-A\) for \(A<0\), tells you where to split the input line. First solve the inequality inside the bars, then substitute the matching signed expression.
For \(h(x)=|x+3|\), the sign changes at \(x=-3\).
\[h(x)=\begin{cases}-(x+3),&x<-3\\x+3,&x\ge-3.\end{cases}\]For \(g(x)=5|x+3|\), multiply both branches by 5: \(-5(x+3)\) on x<−3 and \(5(x+3)\) on x≥−3.
Let \(f(x)=|x^2-4|+1\). The inside is negative when \(-2<x<2\) and nonnegative outside that interval. Therefore
\[f(x)=\begin{cases}x^2-3,&x\le-2\\5-x^2,&-2<x<2\\x^2-3,&x\ge2.\end{cases}\]Check at x=0: \(|0-4|+1=5\), matching the middle branch.
26: split \(|x+3|\) at \(-3\). 27: split \(|x^2-4|+1\) at \(-2\) and 2. 28: split \(5|x+3|\) at \(-3\). 29: split \(|x^2-16|\) at \(-4\) and 4. For item 29 the middle rule is \(16-x^2\), and the outside rule is \(x^2-16\); include the boundary values in either adjoining equal-valued branch, but never give the same input conflicting outputs.
Check your work — Go 21–29 answer key
- \(f^{-1}(x)=\sqrt{x}\) for \(x\ge0\), after restricting the original domain to \([0,\infty)\).
- \(f^{-1}(x)=\frac{x-4}{2}\).
- \(f^{-1}(x)=\sqrt{x}-1\) for \(x\ge0\), after restricting the original domain to \([-1,\infty)\).
- \(h^{-1}(x)=3x-18\).
- \(f^{-1}=\{(5,-3),(-9,-2),(-2,-1),(-5,0),(-4,1),(6,2),(10,3),(8,4)\}\)
- \(h(x)=\begin{cases}-x-3,&x<-3\\x+3,&x\ge-3\end{cases}\)
- \(f(x)=\begin{cases}x^2-3,&x\le-2\\5-x^2,&-2<x<2\\x^2-3,&x\ge2\end{cases}\)
- \(g(x)=\begin{cases}-5x-15,&x<-3\\5x+15,&x\ge-3\end{cases}\)
- \(f(x)=\begin{cases}x^2-16,&x\le-4\\16-x^2,&-4<x<4\\x^2-16,&x\ge4\end{cases}\)
5. Teacher’s review
Part I, graphs 1–4: Endpoints and continuation are not fully identified by arrows or dots, despite requests for exact domain, range, and absolute extrema. Confirm whether students should infer standard full parent-function domains or restrict to the portions shown. A clearer version would use arrows or filled/open endpoints.
Ready 7: Ray AB is said to bisect angle XYZ without making A the vertex Y. State \(A=Y\) as the same plotted point or rename the ray YB.
Go 21 and 23: Quadratic inverses require restricting original domains; the instruction does correctly request restrictions when necessary. Accept either monotone branch when justified.
6. Coverage checklist
| Packet items | What is taught |
|---|---|
| Main task Part I 1–4 | Square-root, exponential, absolute-value and piecewise graph features, intercepts, inverse reflections |
| Main task Part II 5–13 | Two constructions for every requested feature, second differences, one-to-one functions |
| Ready 1–7 | Triangle, ray, line, angle, perpendicular, segment bisector, angle bisector |
| Set 8–20 | Graph features, absolute value, symmetry, minimum/maximum, intercepts, sequences and equations |
| Go 21–25 | Inverse algebra, quadratic domain restrictions, swapping ordered pairs |
| Go 26–29 | Absolute-value piecewise linear and quadratic definitions; two original inverse problems and one ordered-pair problem |
This independently written lesson supports MVP Secondary Math II, Module 4, Section 4.7. Internet access is needed for MathJax equation typesetting; graphs are embedded in the HTML.
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