Free Math Worksheets

- A function assigns exactly one output to each input. - Repeated outputs are allowed. - Look for an input that appears with two different outputs.

1. a) Yes. Each input in \(\{1,2,3,4\}\) is paired with exactly one output. 2. b) Yes. Each input has exactly one output. Different inputs may share the same output without violating the definition of a function. 3. c) No. The input \(5\) is paired with both \(2\) and \(4\).

a) Yes; it is a function. b) Yes; it is a function. c) No; \(x=5\) has two different outputs.

- A function must assign exactly one output to each input. - Decide whether one input could have several outputs. - Imagine entering each assignment as one row in a table.

1. a) This is a function because each date has one recorded daily-high value at the fixed station. 2. b) This is not necessarily a function because one owner may have several registered vehicles and therefore several license plates. 3. c) This is a function because each US state has exactly one state capital.

a) Function b) Not a function c) Function

- Substitute each distance for \(x\). - The base fare stays the same for every trip. - Multiply before adding.

1. For \(x=5\), \(T(5)=3.50+1.80\cdot5=3.50+9.00=12.50\). 2. For \(x=12\), \(T(12)=3.50+1.80\cdot12=3.50+21.60=25.10\). 3. For \(x=0.5\), \(T(0.5)=3.50+1.80\cdot0.5=3.50+0.90=4.40\).

a) \(\$12.50\) b) \(\$25.10\) c) \(\$4.40\)

- Substitute each input for \(x\). - Multiply before subtracting. - Compare your results with the completed entries to check the pattern.

1. For \(x=1\), \(T(1)=7\cdot1-5=2\). 2. For \(x=5\), \(T(5)=7\cdot5-5=30\). 3. For \(x=20\), \(T(20)=7\cdot20-5=135\).

The missing values are \(2\) for \(x=1\), \(30\) for \(x=5\), and \(135\) for \(x=20\).

- Use as many three-scoop deals as possible, then compare with other combinations. - Check whether any fractional number of scoops is a valid input.

1. For \(7\) scoops, buy two three-scoop cups and one single scoop: \(2(\$4.00)+\$1.50=\$9.50\). Seven single scoops would cost \(7(\$1.50)=\$10.50\). 2. The lowest prices are: <table><tr><td>Number of scoops</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td><td>\(6\)</td><td>\(7\)</td><td>\(8\)</td></tr><tr><td>Lowest price</td><td>\(\$1.50\)</td><td>\(\$3.00\)</td><td>\(\$4.00\)</td><td>\(\$5.50\)</td><td>\(\$7.00\)</td><td>\(\$8.00\)</td><td>\(\$9.50\)</td><td>\(\$11.00\)</td></tr></table> 3. The domain consists of whole numbers because the shop sells whole scoops. Values between consecutive whole numbers do not represent possible purchases, so the graph would contain separate points rather than connected line segments.

a) Two three-scoop cups and one single scoop, costing \(\$9.50\) b) <table><tr><td>Number of scoops</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td><td>\(6\)</td><td>\(7\)</td><td>\(8\)</td></tr><tr><td>Lowest price</td><td>\(\$1.50\)</td><td>\(\$3.00\)</td><td>\(\$4.00\)</td><td>\(\$5.50\)</td><td>\(\$7.00\)</td><td>\(\$8.00\)</td><td>\(\$9.50\)</td><td>\(\$11.00\)</td></tr></table> c) The relation is discrete because only whole numbers of scoops can be purchased.

- Identify the lightest and heaviest possible combinations. - Find how much the weight changes when one workbook is replaced by one textbook. - Decide whether a fractional number of textbooks is possible.

1. The minimum occurs with \(6\) workbooks: \(6(0.25)=1.5\,\text{lb}\). The maximum occurs with \(6\) textbooks: \(6(1.5)=9\,\text{lb}\). 2. Replacing one workbook with one textbook increases the total weight by \(1.5-0.25=1.25\,\text{lb}\). The completed table is: <table><tr><td>Number of textbooks</td><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td><td>\(6\)</td></tr><tr><td>Total weight (lb)</td><td>\(1.5\)</td><td>\(2.75\)</td><td>\(4\)</td><td>\(5.25\)</td><td>\(6.5\)</td><td>\(7.75\)</td><td>\(9\)</td></tr></table> 3. The number of textbooks must be a whole number. Inputs such as \(1.5\) textbooks have no meaning, so the graph is discrete and the points should not be connected.

a) Minimum: \(1.5\,\text{lb}\); maximum: \(9\,\text{lb}\) b) <table><tr><td>Number of textbooks</td><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td><td>\(6\)</td></tr><tr><td>Total weight (lb)</td><td>\(1.5\)</td><td>\(2.75\)</td><td>\(4\)</td><td>\(5.25\)</td><td>\(6.5\)</td><td>\(7.75\)</td><td>\(9\)</td></tr></table> c) No. Only whole-number inputs are meaningful.

- Substitute the given input into each function. - Evaluate inside parentheses first. - Be careful when multiplying two negative numbers.

1. \(T_1(-0.75)=4\cdot(-0.75+1.5)=4\cdot0.75=3\). 2. \(T_2(-0.75)=2-6\cdot(-0.75)=2+4.5=6.5\). 3. Since \(6.5>3\), \(T_2\) has the greater value.

\(T_1(-0.75)=3\) and \(T_2(-0.75)=6.5\). Therefore, \(T_2\) has the greater value.

- Substitute each input for \(x\). - Multiply before subtracting. - Simplify fractions during the calculation. - Equal quantities have a difference of \(0\).

1. \(T(3)=\frac{2}{3}(3)-\frac{1}{2}=2-\frac{1}{2}=\frac{3}{2}\). 2. \(T\left(\frac{3}{4}\right)=\frac{2}{3}\cdot\frac{3}{4}-\frac{1}{2}=\frac{1}{2}-\frac{1}{2}=0\). 3. \(T(6)=\frac{2}{3}(6)-\frac{1}{2}=4-\frac{1}{2}=\frac{7}{2}\). 4. \(T\left(\frac{1}{4}\right)=\frac{2}{3}\cdot\frac{1}{4}-\frac{1}{2}=\frac{1}{6}-\frac{3}{6}=-\frac{1}{3}\).

a) \(\frac{3}{2}\) b) \(0\) c) \(\frac{7}{2}\) d) \(-\frac{1}{3}\)

- Substitute the input for \(x\). - Use parentheses when substituting a negative number. - Convert \(\frac{1}{2}\) to a decimal if helpful.

1. \(T(8.9)=-15.4+8.9=-6.5\). 2. \(T(-3.7)=-15.4+(-3.7)=-19.1\). 3. \(T(15.4)=-15.4+15.4=0\). 4. Since \(\frac{1}{2}=0.5\), \(T\left(\frac{1}{2}\right)=-15.4+0.5=-14.9\).

a) \(-6.5\) b) \(-19.1\) c) \(0\) d) \(-14.9\)

- Apply the definition of a function to the time values. - Compare the differences between consecutive temperatures. - Subtract the minimum temperature from the maximum temperature.

1. Each recorded time is paired with exactly one temperature, so the data define a function. 2. From \(3{:}00\) p.m. to \(6{:}00\) p.m., the change is \(39-46=-7\,^\circ\text{F}\). From \(6{:}00\) p.m. to \(9{:}00\) p.m., the change is \(31-39=-8\,^\circ\text{F}\). The greatest decrease occurred from \(6{:}00\) p.m. to \(9{:}00\) p.m. 3. The highest temperature is \(46\,^\circ\text{F}\) and the lowest is \(28\,^\circ\text{F}\). The range is \(46-28=18\,^\circ\text{F}\).

a) Each time has exactly one recorded temperature. b) From \(6{:}00\) p.m. to \(9{:}00\) p.m.; the temperature decreased by \(8\,^\circ\text{F}\). c) \(18\,^\circ\text{F}\)

- In a list of ordered pairs, repeated x-values with different y-values violate the function rule. - Change one of the two points whose x-coordinate is \(2\). - To test the reverse assignment, switch the roles of \(x\) and \(y\).

1. The input \(x=2\) occurs in \(P_2(2, 6)\) and \(P_4(2, 1)\) with two different outputs. Therefore, \(M\) does not represent a function. 2. One possible change is to replace \(P_4(2, 1)\) with \(P_4(4, 1)\). The new x-values are \(1\), \(2\), \(3\), and \(4\), so each input has exactly one output. 3. In the new set, the output \(y=3\) occurs at both \(x=1\) and \(x=3\). Therefore, the reverse assignment is not a function.

a) \(x=2\) is paired with both \(y=6\) and \(y=1\). b) One answer is \(P_4(4, 1)\). c) No. The input \(y=3\) in the reverse assignment would have outputs \(x=1\) and \(x=3\).

- Substitute a point's x-coordinate into the equation and compare the output with its y-coordinate. - A point lies on the line when its coordinates make the equation true. - If the y-coordinate is known, substitute it and solve for the unknown x-coordinate.

1. For \(A(-2, 6)\), substitute \(x = -2\): \(y = -1.5 \cdot (-2) + 3 = 6\). Therefore, \(A\) lies on the line. 2. For \(B(4, -3)\), substitute \(x = 4\): \(y = -1.5 \cdot 4 + 3 = -3\). Therefore, \(B\) lies on the line. 3. For \(C(k, 0)\), set \(y = 0\): \(0 = -1.5k + 3\). Solving gives \(k = 2\).

a) Yes, both \(A\) and \(B\) lie on the line. b) \(k = 2\)

- Write the coordinate that stays constant on each axis. - A function may assign only one y-value to each x-value. - Apply the vertical line test.

1. The x-axis has equation \(y=0\). It is the graph of the linear function \(f(x)=0\), because every real input has exactly one output, \(0\). 2. The y-axis has equation \(x=0\). It is not the graph of \(y=f(x)\) because the input \(x=0\) is paired with infinitely many y-values. Equivalently, it fails the vertical line test.

a) \(y=0\); yes, it is the graph of a linear function. b) \(x=0\); no, it is not the graph of \(y=f(x)\).

- For each x-coordinate, calculate the y-value of the line. - Compare that value with the point's y-coordinate. - What does a larger or smaller y-coordinate mean at the same x-value?

1. At \(x = 2\), the line has \(y = 1.2 \cdot 2 + 0.5 = 2.9\). Since this matches the y-coordinate of \(P\), \(P\) is on the line. 2. At \(x = -1\), the line has \(y = 1.2(-1) + 0.5 = -0.7\). Since \(-1 < -0.7\), \(Q\) is below the line. 3. At \(x = 5\), the line has \(y = 1.2 \cdot 5 + 0.5 = 6.5\). Since this matches the y-coordinate of \(R\), \(R\) is on the line.

\(P\) is on the line. \(Q\) is below the line. \(R\) is on the line.

- Substitute each input for \(x\). - Use parentheses around negative inputs. - Work with either fractions or decimals consistently.

1. \(T(3)=-4\cdot3+\frac{1}{2}=-12+0.5=-11.5\). 2. \(T(-1.5)=-4\cdot(-1.5)+0.5=6.5\). 3. \(T\left(\frac{3}{8}\right)=-4\cdot\frac{3}{8}+\frac{1}{2}=-\frac{3}{2}+\frac{1}{2}=-1\).

a) \(-11.5\) b) \(6.5\) c) \(-1\)

- Substitute the number of miles for \(s\). - Multiply before adding. - Interpret each output as a cost in dollars.

1. For \(s=80\), \(K=42+0.15\cdot80=42+12=54\). 2. For \(s=240\), \(K=42+0.15\cdot240=42+36=78\).

1) \(\$54.00\) 2) \(\$78.00\)

- Start at each time on the horizontal axis and trace to the graph, then read the corresponding temperature. - To find the increase, subtract the starting temperature from the ending temperature. - Pay attention to the signs of both temperatures. - For part c), find where the graph crosses the horizontal level \(-8\).

1. At \(2{:}00\) a.m., the graph has an output of \(-12\,^\circ\text{C}\). At \(2{:}00\) p.m., which is \(14\) hours after midnight, the output is \(-4\,^\circ\text{C}\). 2. The temperature increase is \(-4-(-12)=8\,^\circ\text{C}\). 3. The graph has an output of \(-8\,^\circ\text{C}\) at \(8\) and \(20\) hours after midnight, which are \(8{:}00\) a.m. and \(8{:}00\) p.m.

a) \(2{:}00\) a.m.: \(-12\,^\circ\text{C}\); \(2{:}00\) p.m.: \(-4\,^\circ\text{C}\) b) \(8\,^\circ\text{C}\) c) \(8{:}00\) a.m. and \(8{:}00\) p.m.

- Recall that a function assigns exactly one output to each input. - Read the graph at \(t = 0, 2, 4, 6, 8, 10,\) and \(12\). - To find when the depth is \(4\,\text{m}\), trace horizontally from \(h = 4\) to the graph. - For the reverse relation, ask whether one depth can occur at two different times.

1. Read the plotted values: <table> <tr><th>Time \(t\) (h)</th><td>\(0\)</td><td>\(2\)</td><td>\(4\)</td><td>\(6\)</td><td>\(8\)</td><td>\(10\)</td><td>\(12\)</td></tr> <tr><th>Depth \(h\) (m)</th><td>\(1\)</td><td>\(2\)</td><td>\(4\)</td><td>\(5\)</td><td>\(4\)</td><td>\(2\)</td><td>\(1\)</td></tr> </table> 2. The relation from time to depth is a function because each time is paired with exactly one depth. 3. The graph has depth \(4\,\text{m}\) at \(t = 4\,\text{h}\) and \(t = 8\,\text{h}\). 4. The reverse relation is not a function because the input depth \(4\,\text{m}\) would be paired with two different times.

a) <table> <tr><th>Time \(t\) (h)</th><td>\(0\)</td><td>\(2\)</td><td>\(4\)</td><td>\(6\)</td><td>\(8\)</td><td>\(10\)</td><td>\(12\)</td></tr> <tr><th>Depth \(h\) (m)</th><td>\(1\)</td><td>\(2\)</td><td>\(4\)</td><td>\(5\)</td><td>\(4\)</td><td>\(2\)</td><td>\(1\)</td></tr> </table> Yes. Time to depth is a function because each time has exactly one depth. b) The depth is \(4\,\text{m}\) at \(t = 4\,\text{h}\) and \(t = 8\,\text{h}\). The reverse relation is not a function because one depth can correspond to more than one time.

- Identify which coordinate stays constant on each line. - A vertical line has a constant x-coordinate; a horizontal line has a constant y-coordinate. - Apply the vertical line test.

1. Line \(g\) is vertical at \(x=3\), so one equation is \(x+0y=3\). 2. Line \(g\) is not the graph of a function \(y=f(x)\) because the input \(x=3\) corresponds to infinitely many y-values. It fails the vertical line test. 3. Line \(h\) is horizontal at \(y=-2\), so one equation is \(0x+y=-2\). 4. Line \(h\) is the graph of the constant function \(f(x)=-2\), because each input has exactly one output.

a) \(g:\ x+0y=3\) \(h:\ 0x+y=-2\) b) \(g\) is not a function graph. \(h\) is a function graph.

- Compare an upper semicircle with a full circle. - Apply the vertical line test.

1. The graph is the upper semicircle \(y=\sqrt{4-x^2}\) for \(-2\le x\le2\). 2. Every input in \([-2, 2]\) corresponds to exactly one output on the upper semicircle. Equivalently, every vertical line meets the graph at most once. 3. Therefore, the graph represents a function.

Yes. The graph represents a function because each \(x\in[-2, 2]\) has exactly one y-value.

- Look for a vertical line that intersects the graph more than once. - Check the graph at \(x=2\).

1. A graph represents a function only if every input \(x\) has at most one output \(y\). 2. The curve is a sideways V. For example, the vertical line \(x=2\) meets the curve at \((2, 3)\) and \((2, 1)\). 3. Therefore, the curve fails the vertical line test and is not the graph of a function.

No. At \(x=2\), the graph has two y-values, \(3\) and \(1\).

- Substitute each point's x-coordinate into the function. - Compare the function output with the point's y-coordinate. - Use the graph to check whether your calculations are reasonable.

1. For \(A(2, 1)\), substitute \(x = 2\): \(1.5 \cdot 2 - 2 = 1\). This matches the point's y-coordinate, so \(A\) lies on the graph. 2. For \(B(-1, -3.5)\), substitute \(x = -1\): \(1.5 \cdot (-1) - 2 = -3.5\). This matches the point's y-coordinate, so \(B\) lies on the graph. 3. For \(C(4, 3.5)\), substitute \(x = 4\): \(1.5 \cdot 4 - 2 = 4\). Since \(4 \neq 3.5\), \(C\) does not lie on the graph.

Points \(A(2, 1)\) and \(B(-1, -3.5)\) lie on the graph. Point \(C(4, 3.5)\) does not.

- Move an imaginary vertical line across each graph. - A function graph may be intersected at most once by any vertical line.

1. Graph a) is a function because every vertical line intersects the parabola at most once. 2. Graph b) is not a function because vertical lines through \(-2<x<2\) intersect the circle twice. 3. Graph c) is a function because every vertical line intersects the line at most once. 4. Graph d) is not a function because vertical lines through \(0<x\le4\) intersect the sideways V twice.

Graphs a) and c) are functions. Graphs b) and d) are not functions.

- Evaluate both functions for each input. - Organize the outputs in matching rows. - Look for a column with equal outputs.

1. The outputs for \(A(x)\) are \(16,18,20,22,\) and \(24\). 2. The outputs for \(B(x)\) are \(12,16,20,24,\) and \(28\). 3. Both functions have output \(20\) when \(x=5\).

a) <table> <tr><th>\(x\)</th><th>\(3\)</th><th>\(4\)</th><th>\(5\)</th><th>\(6\)</th><th>\(7\)</th></tr> <tr><th>\(A(x)\)</th><td>\(16\)</td><td>\(18\)</td><td>\(20\)</td><td>\(22\)</td><td>\(24\)</td></tr> <tr><th>\(B(x)\)</th><td>\(12\)</td><td>\(16\)</td><td>\(20\)</td><td>\(24\)</td><td>\(28\)</td></tr> </table> b) \(x=5\); both outputs are \(20\).

- Substitute the given input for \(x\). - Evaluate the expression inside parentheses before multiplying. - Simplify fractions before multiplying when possible. - Check the sign of the final result.

1. Substitute \(x=\frac{3}{4}\): \(T\left(\frac{3}{4}\right)=\frac{2}{3}\cdot\frac{3}{4}-4\cdot\left(\frac{3}{4}+\frac{1}{2}\right)\). 2. The first term is \(\frac{2}{3}\cdot\frac{3}{4}=\frac{1}{2}\). 3. Inside the parentheses, \(\frac{3}{4}+\frac{1}{2}=\frac{5}{4}\). 4. Then \(4\cdot\frac{5}{4}=5\). 5. Therefore, \(T\left(\frac{3}{4}\right)=\frac{1}{2}-5=-\frac{9}{2}=-4.5\).

\(-\frac{9}{2}\), or \(-4.5\)

- Substitute each given input into both functions. - Compare the outputs at the smaller and larger inputs. - Test integer values between \(2\) and \(10\).

1. For \(k=2\), \(T_1(2)=2\cdot2+10=14\) and \(T_2(2)=40-3\cdot2=34\). 2. For \(k=10\), \(T_1(10)=2\cdot10+10=30\) and \(T_2(10)=40-3\cdot10=10\). 3. \(T_1\) increases as \(k\) increases, while \(T_2\) decreases. 4. Testing \(k=6\) gives \(T_1(6)=22\) and \(T_2(6)=22\), so the outputs are equal.

a) For \(k=2\): \(T_1=14\), \(T_2=34\). For \(k=10\): \(T_1=30\), \(T_2=10\). b) \(T_1\) increases and \(T_2\) decreases. c) \(k=6\); both outputs are \(22\).

- Substitute the number of minutes for \(m\). - Compare the two outputs for part b). - In part c), set \(0.20m\) equal to \(5\).

1. For \(m=50\), \(K_A=5+0.12\cdot50=11\) and \(K_B=0.20\cdot50=10\). 2. For \(m=100\), \(K_A=5+0.12\cdot100=17\) and \(K_B=0.20\cdot100=20\). Plan A costs less. 3. Set Plan B's cost equal to \(5\): \(0.20m=5\). Then \(m=5\div 0.20=25\).

a) Plan A: \(\$11.00\); Plan B: \(\$10.00\) b) Plan A; it costs \(\$17.00\), compared with \(\$20.00\) for Plan B. c) \(25\) minutes

- Substitute the given input into both functions. - Equal function values must have identical outputs. - Follow the order of operations when evaluating.

1. \(T_1(3)=24-4\cdot3=12\) and \(T_2(3)=2\cdot3+6=12\). 2. \(T_1(5)=24-4\cdot5=4\) and \(T_2(5)=2\cdot5+6=16\). Since \(4\ne 16\), the claim is false. 3. \(T_1(10)=24-4\cdot10=-16\).

a) \(T_1(3)=12\) and \(T_2(3)=12\) b) The claim is false because \(4\ne 16\). c) \(T_1(10)=-16\)

- Look for an interval in which the battery level increases. - Divide the total decrease by the elapsed time. - For the reverse assignment, check whether one battery percentage appears at more than one time.

1. The battery level rises from \(44\%\) at \(4{:}00\) p.m. to \(88\%\) at \(6{:}00\) p.m., so the phone was probably charging during that interval. 2. From \(7{:}00\) a.m. to \(12{:}00\) p.m. is \(5\) hours. The battery decreases by \(100-60=40\) percentage points, so the average use is \(40\div5=8\) percentage points per hour. 3. The reverse assignment is not a function because the battery level \(60\%\) corresponds to both \(12{:}00\) p.m. and \(2{:}00\) p.m.

a) From \(4{:}00\) p.m. to \(6{:}00\) p.m. b) \(8\) percentage points per hour c) No. The input \(60\%\) would have two outputs: \(12{:}00\) p.m. and \(2{:}00\) p.m.

- Substitute the given x-value into the function to find its output. - Moving vertically changes the y-coordinate but not the x-coordinate. - To test a point, evaluate the function at the point's x-coordinate and compare the result with its y-coordinate.

1. \(f(4) = 1.5 \cdot 4 - 2 = 4\), so \(P(4, 4)\). 2. Moving \(5\) units vertically upward keeps the x-coordinate the same and adds \(5\) to the y-coordinate. Thus \(Q(4, 9)\). 3. \(g(4) = 1.5 \cdot 4 + 3 = 9\). This matches the y-coordinate of \(Q\), so \(Q\) lies on the graph of \(g\).

1) \(y_P = 4\) 2) \(Q(4, 9)\) 3) Yes, because \(g(4) = 9\).

- Find a number in \(D\) with more than one distinct prime factor. - The revised rule must produce exactly one output even for \(n=1\). - Ask how many absolute values one rational number has.

1. The input \(6\) has the distinct prime factors \(2\) and \(3\), so it would receive two different outputs. Therefore, the rule does not define a function. 2. One valid revision is: “Assign each \(n\) the number of its distinct prime factors.” This rule assigns exactly one output to every input, including assigning \(0\) to \(1\). 3. Every rational number has exactly one absolute value, so \(x\mapsto|x|\) is a function.

a) No. For example, \(6\) would have outputs \(2\) and \(3\). b) Example: Assign each \(n\) the number of its distinct prime factors, with \(1\mapsto0\). c) Yes. Every rational number has exactly one absolute value.

- Pay close attention to which endpoints are included in each interval. - A function assigns exactly one output to every input in its domain. - In the reverse direction, ask whether a postage value determines one exact weight.

1. Apply the rate interval that contains each weight. The postage values are \(\$0.85\), \(\$0.85\), \(\$1.00\), \(\$1.00\), and \(\$1.60\). 2. Each weight in \(0<m\le16\) belongs to exactly one rate interval, so it has exactly one postage value. Therefore, weight \(\rightarrow\) postage is a function. 3. The reverse assignment is not a function. For example, the postage value \(\$1.00\) corresponds to every weight in \(1<m\le2\), not to one unique weight.

a) <table><tr><td>Weight (oz)</td><td>\(0.5\)</td><td>\(1\)</td><td>\(1.5\)</td><td>\(2\)</td><td>\(5\)</td></tr><tr><td>Postage</td><td>\(\$0.85\)</td><td>\(\$0.85\)</td><td>\(\$1.00\)</td><td>\(\$1.00\)</td><td>\(\$1.60\)</td></tr></table> b) Yes. Each allowed weight has exactly one postage value. c) No. One postage value corresponds to many possible weights.

- Compare the x-coordinates of the four points. - One input cannot have several outputs in a function. - After switching the coordinates, identify the new input set and the common output.

1. The points \((4, -2)\), \((4, 0)\), \((4, 2)\), and \((4, 5)\) all have x-coordinate \(4\), so they lie on the vertical line \(x=4\). 2. The input \(x=4\) is assigned four different outputs, so \(x\rightarrow y\) is not a function. 3. After switching the coordinates, the inputs are \(-2\), \(0\), \(2\), and \(5\), and every output is \(4\). Thus, \(f(u)=4\) for \(u\in\{-2, 0, 2, 5\}\).

a) The points lie on \(x=4\). b) It is not a function because \(x=4\) has multiple outputs. c) \(f(u)=4\) for \(u\in\{-2, 0, 2, 5\}\).

- Use distance divided by time for each speed. - Average speed for the whole trip uses total distance divided by total time. - In part c), only the return time varies.

1. The speed to the shelter is \(v_1 = \frac{6}{1.5} = 4\,\text{mph}\). 2. The return time is \(t_2 = \frac{9}{4.5} = 2\) hours. 3. The total distance is \(15\) miles, and the total time is \(1.5 + 2 = 3.5\) hours. 4. The average speed for the entire hike is \(v_g = \frac{15}{3.5} \approx 4.29\,\text{mph}\). 5. If the return time is variable, the total time is \(1.5 + t_2\), so \(v_g(t_2) = \frac{15}{1.5 + t_2}\).

a) \(v_1 = 4\,\text{mph}\) b) \(t_2 = 2\) hours, and \(v_g \approx 4.29\,\text{mph}\) c) \(v_g(t_2) = \frac{15}{1.5 + t_2}\)

- First determine the total time allowed by the desired round-trip average. - Subtract the fixed outbound time to find the return time. - In part c), interpret what a return time of zero would mean.

1. The trip to work takes \(0.5\) hour, so \(v_1 = \frac{20}{0.5} = 40\,\text{mph}\). 2. Let \(t_2\) be the return time in hours. For a \(50\,\text{mph}\) round-trip average, write \(50 = \frac{40}{0.5 + t_2}\). 3. Solve: \(0.5 + t_2 = \frac{40}{50} = 0.8\), so \(t_2 = 0.3\) hour. 4. The required return speed is \(v_2 = \frac{20}{0.3} = \frac{200}{3}\,\text{mph} \approx 66.67\,\text{mph}\). 5. For an \(80\,\text{mph}\) round-trip average, \(80 = \frac{40}{0.5 + t_2}\) gives \(0.5 + t_2 = 0.5\), so \(t_2 = 0\). 6. A zero return time would require infinite speed, so an \(80\,\text{mph}\) round-trip average is not attainable.

a) \(v_1 = 40\,\text{mph}\) b) \(v_2 = \frac{200}{3}\,\text{mph} \approx 66.67\,\text{mph}\) c) No. It would require a return time of \(0\) hours.

- A negative number squared is positive. - Evaluate both functions for each listed input. - Compare the two outputs in each row.

1. For \(y=-4\), the outputs are \(-5\) and \(12\). 2. For \(y=-2\), the outputs are \(-4\) and \(0\). 3. For \(y=0\), the outputs are \(-3\) and \(-4\). 4. For \(y=2\), the outputs are \(-2\) and \(0\). 5. For \(y=4\), the outputs are \(-1\) and \(12\). 6. Only at \(y=0\) is \(T_1(y)>T_2(y)\).

<table> <tr><th>\(y\)</th><th>\(T_1(y)\)</th><th>\(T_2(y)\)</th></tr> <tr><td>\(-4\)</td><td>\(-5\)</td><td>\(12\)</td></tr> <tr><td>\(-2\)</td><td>\(-4\)</td><td>\(0\)</td></tr> <tr><td>\(0\)</td><td>\(-3\)</td><td>\(-4\)</td></tr> <tr><td>\(2\)</td><td>\(-2\)</td><td>\(0\)</td></tr> <tr><td>\(4\)</td><td>\(-1\)</td><td>\(12\)</td></tr> </table> \(T_1(y)>T_2(y)\) only when \(y=0\).

- Evaluate each expression separately. - Apply exponents before multiplication and subtraction. - Subtract the smaller output from the larger output.

1. For \(y=3\), \(T_1=5\cdot3-15=0\) and \(T_2=3^2-3\cdot3=0\). 2. For \(y=6\), \(T_1=5\cdot6-15=15\) and \(T_2=6^2-3\cdot6=18\). 3. At \(y=6\), \(T_2\) is greater by \(18-15=3\).

a) \(T_1=0\) and \(T_2=0\) b) \(T_1=15\) and \(T_2=18\) c) \(T_2\) is greater by \(3\).

- Substitute each input into the function rule. - Represent a required output with an equation. - Set two function rules equal when their outputs must match. - Use equivalent operations to isolate the input.

1. Evaluate \(T(x) = 5x - 7\) at each input: \(T(-3) = 5 \cdot (-3) - 7 = -22\), \(T(0) = 5 \cdot 0 - 7 = -7\), and \(T(2.5) = 5 \cdot 2.5 - 7 = 5.5\). 2. Solve \(5x - 7 = 13\). Add \(7\): \(5x = 20\). Divide by \(5\): \(x = 4\). 3. Set the outputs equal: \(5x - 7 = 2x + 8\). Subtract \(2x\) and add \(7\): \(3x = 15\). Divide by \(3\): \(x = 5\).

1) The missing outputs are \(-22\), \(-7\), and \(5.5\). 2) \(x = 4\) 3) \(x = 5\)

- Substitute each value of \(m\) into both formulas. - The constant term is the cost at \(0\) minutes. - Compare the two entries in each column.

1. Plan A costs \(\$10.00\), \(\$12.50\), \(\$15.00\), and \(\$17.50\) for the listed inputs. 2. Plan B costs \(\$5.00\), \(\$10.00\), \(\$15.00\), and \(\$20.00\). 3. The plans cost the same at \(100\) minutes. At \(150\) minutes, Plan A is less expensive.

a) <table> <tr><td>Minutes \((m)\)</td><td>\(0\)</td><td>\(50\)</td><td>\(100\)</td><td>\(150\)</td></tr> <tr><td>Plan A</td><td>\(\$10.00\)</td><td>\(\$12.50\)</td><td>\(\$15.00\)</td><td>\(\$17.50\)</td></tr> <tr><td>Plan B</td><td>\(\$5.00\)</td><td>\(\$10.00\)</td><td>\(\$15.00\)</td><td>\(\$20.00\)</td></tr> </table> b) At \(150\) minutes, Plan A is less expensive: \(\$17.50<\$20.00\).

- Solve each equation for \(y\) when possible and compare the resulting shape with the graphs. - Use the vertical line test to decide whether each relation is a function. - A proportional linear function has the form \(y=kx\) and passes through the origin.

1. Graph 1 is Equation A. Solving \(2y-3x=0\) for \(y\) gives \(y=\frac{3}{2}x\), a line through the origin with slope \(\frac{3}{2}\). 2. Graph 2 is Equation B. The relation \(y^2=x\) is \(y=\pm\sqrt{x}\), a sideways parabola. 3. Graph 3 is Equation C. Solving \(x^2+y=4\) gives \(y=4-x^2\), a downward-opening parabola with vertex \((0,4)\). 4. Graph 1 is a function because each \(x\)-value has exactly one \(y\)-value. Graph 2 is not a function because every \(x>0\) has two \(y\)-values. Graph 3 is a function because each \(x\)-value has exactly one \(y\)-value. 5. For Graph 1, \(y=\frac{3}{2}x\), so it is proportional. For Graph 3, \(y=4-x^2\), so it is neither linear nor proportional.

a) A \(\rightarrow\) Graph 1 B \(\rightarrow\) Graph 2 C \(\rightarrow\) Graph 3 b) Graph 1: function Graph 2: not a function Graph 3: function c) Graph 1: \(y=\frac{3}{2}x\); proportional Graph 3: \(y=4-x^2\); neither linear nor proportional

- Interpret increasing, horizontal, and vertical segments in the context. - Time cannot reverse direction in a realistic time graph. - Use the vertical line test.

1. Graph A is realistic. Increasing segments represent walking away from home, and horizontal segments represent stopping, such as at a crosswalk. 2. Graph B is not realistic. At \(t=2\) minutes, it contains a vertical segment, which would place the student at many distances at the same instant. It also moves backward in time from \(t=6\) to \(t=5\), which cannot represent a time-ordered trip. 3. Graph A passes the vertical line test, so it is a function graph. 4. Graph B fails the vertical line test. For example, \(t=2\) has many distances, and some inputs between \(5\) and \(6\) minutes have multiple distances.

a) Graph A is realistic. Graph B is impossible because it includes an instantaneous change in distance and a segment that moves backward in time. b) Graph A is a function graph. Graph B is not a function graph.

- The maximum is the graph's highest point. - Read the graph at the requested time values. - For the reverse relation, check whether one height can be paired with more than one time.

1. The vertex of the graph is \((3, 18)\), so the ball reaches a maximum height of \(18\,\text{ft}\) after \(3\) seconds. 2. The graph gives \(h(1)=10\) and \(h(5)=10\), so the ball is \(10\,\text{ft}\) high at both times. 3. The reverse relation is not a function because most heights correspond to two different times: once while the ball is rising and once while it is falling. For example, a height of \(10\,\text{ft}\) corresponds to both \(t=1\) and \(t=5\).

a) \(18\,\text{ft}\) after \(3\) seconds b) \(10\,\text{ft}\) at both \(1\) second and \(5\) seconds c) No. A single height can correspond to two different times.

- Apply the vertical line test to an x-value near the center of the circle. - Consider splitting the circle horizontally rather than vertically.

1. For most x-values between the leftmost and rightmost points of the circle, a vertical line meets the shoreline twice: once on the upper half and once on the lower half. Therefore, one input would have two outputs, so the entire circle is not the graph of \(y=f(x)\). 2. Divide the circle horizontally into an upper semicircle and a lower semicircle. Each semicircle passes the vertical line test and can be represented as a separate function.

a) The circle is not a function graph because many x-values correspond to two y-values. b) Split it into the upper semicircle and the lower semicircle.

- The trip ends when the graph returns to distance \(0\). - A horizontal segment means the distance from the center is unchanged. - Use the constant rate on the final segment to find the value at \(45\) minutes. - Apply the definition of a function or the vertical line test.

1. The graph begins at \(t=0\) and returns to distance \(0\) at \(t=60\) minutes, so the trip lasts \(60\) minutes. 2. At \(t=20\), the graph is on a horizontal segment at \(4\) miles. The courier is probably making a delivery or taking a short break. 3. From minute \(40\) to minute \(60\), the distance decreases linearly from \(6\) miles to \(0\). In \(5\) minutes, it decreases by \(\frac{5}{20}\cdot6=1.5\) miles, so after \(45\) minutes the distance is \(6-1.5=4.5\) miles. 4. Every input time is paired with exactly one distance. Equivalently, every vertical line intersects the graph at most once, so the relation is a function.

a) \(60\) minutes b) \(4\) miles; probably making a delivery or taking a break c) \(4.5\) miles d) Each time corresponds to exactly one distance.

- Locate the graph's highest point. - Compare the graph with a horizontal line at \(15\). - Check the early-morning portion of the graph for part c). - For the reverse relation, ask whether one data-rate value can occur at more than one time.

1. The graph reaches its maximum at about \(8{:}00\) p.m., with a data rate of about \(42\,\text{Mbps}\). 2. The graph is above \(15\,\text{Mbps}\) from about noon to \(2{:}30\) p.m. and from about \(4{:}30\) p.m. to \(11{:}00\) p.m. 3. The claim is false. During the overnight and early-morning hours, the graph is below \(5\,\text{Mbps}\). 4. The reverse relation is not a function because the same data rate occurs at several different times during the day. One input data rate would be paired with more than one output time.

a) About \(8{:}00\) p.m.; about \(42\,\text{Mbps}\) b) About noon–\(2{:}30\) p.m. and \(4{:}30\) p.m.–\(11{:}00\) p.m. c) False; the rate falls below \(5\,\text{Mbps}\) overnight. d) No; one data rate can correspond to several times.

- “Started hour” means that one minute into a new hour triggers the next charge. - Ask whether one fee can occur for two different parking times. - A reverse function requires the original pricing rule to be one-to-one.

1. A \(30\)-minute or \(60\)-minute stay is within the first started hour, so each costs \(\$1.50\). A \(61\)-minute stay starts a second hour, so it costs \(\$1.50+\$1.00=\$2.50\). A \(120\)-minute stay also costs \(\$2.50\). 2. The reverse assignment is not a function because one fee can correspond to many parking times. For example, \(\$1.50\) corresponds to both \(30\) minutes and \(60\) minutes. 3. A strictly increasing rule such as \(P(t)=kt\), where \(k>0\) and \(t\) is the exact parking time, assigns different fees to different times. Its reverse therefore defines a function.

a) \(30\) minutes: \(\$1.50\) \(60\) minutes: \(\$1.50\) \(61\) minutes: \(\$2.50\) \(120\) minutes: \(\$2.50\) b) The reverse assignment is not a function because one fee corresponds to multiple parking times. c) One possible rule is \(P(t)=kt\) with \(k>0\), using exact time and no rounding.