- Identify the highest point on the graphed path. - Read both coordinates of that point and match each coordinate to its axis label.

1. The maximum occurs at the highest point of the parabola. 2. The graph's highest point is \((3, 9)\), so the rocket reaches a maximum height of \(9\,\text{ft}\) at \(3\,\text{s}\).

The maximum height is \(9\,\text{ft}\), and it occurs at \(3\,\text{s}\).

- Identify the vertex directly from the quadratic's vertex form. - Interpret the vertex's x-coordinate using the stated unit for \(x\), and interpret its y-coordinate using the stated unit for \(P(x)\).

1. The model is in vertex form, so its vertex is \((6, 7200)\). 2. Because the coefficient of the squared term is negative, the vertex gives the maximum. 3. Since \(x=6\) represents \(6\) hundreds of units, the maximizing sales level is \(600\) units, and the maximum weekly profit is \(\$7200\).

Sell \(600\) units for a maximum weekly profit of \(\$7200\).

- What time does the input \(t=0\) stand for? - Substitute that time into the function. - Interpret the constant term as the graph's y-intercept in this context.

1. The input \(t=0\) is the beginning of the shift. 2. Substituting gives \(C(0)=0.5\cdot0^2+4\cdot0+30=30\). 3. The model gives an initial operating cost of \(\$30\).

\(C(0)=\$30\), the modeled cost at the beginning of the shift.

- What does an x-intercept mean for the output value of a function? - Translate an output of \(0\) into the meaning of profit. - Convert the graph's x-values from hundreds of units to units.

1. The x-intercepts occur where profit is \(0\). 2. From the graph, the intercepts are at \(x=2\) and \(x=8\). 3. Because \(x\) is measured in hundreds of units, the company breaks even at \(200\) units and \(800\) units.

The x-intercepts are break-even sales levels: \(200\) units and \(800\) units.

- Use only x-values that correspond to points on the physical arch. - Identify the smallest and largest heights attained on that interval. - The vertex gives one endpoint of the range.

1. The physical arch runs from the left base at \(x=0\) to the right base at \(x=8\), so the contextual domain is \([0,8]\). 2. The arch is at ground level at the endpoints, so the minimum modeled height is \(0\). 3. The vertex is \((4,8)\), so the maximum modeled height is \(8\). 4. Therefore, the contextual range is \([0,8]\) feet.

Domain: \([0,8]\) Range: \([0,8]\) feet

- Compare how much \(y\) changes each time \(x\) increases by \(1\). - Do those changes stay constant? - If the first changes are not constant, compare how those changes themselves vary.

1. The first differences are \(3\), \(5\), and \(7\), so they are not constant. 2. The differences between those first differences are \(2\) and \(2\), so the second differences are constant. 3. Constant second differences are consistent with a quadratic model.

A quadratic model is more consistent. The first differences are \(3,5,7\), and the second differences are constant at \(2\).

- Which quantity does the input represent? - Substitute the stated time into the model. - Give the output with its distance unit.

1. The input \(t=3\) means \(3\,\text{s}\) after the car starts rolling. 2. Substitute: \(d(3)=2\cdot3^2=18\). 3. The car's modeled distance traveled in \(3\,\text{s}\) is \(18\,\text{ft}\).

\(d(3)=18\,\text{ft}\), the modeled distance traveled after \(3\,\text{s}\).

- What do the two factors in the quadratic area rule represent? - At what width would one of the two sides shrink to zero? - Check that both factors stay positive throughout your proposed interval.

1. The quadratic area rule is a product of the two side lengths, \(w\) and \(12-w\). 2. Both factors must be positive: \(w>0\) and \(12-w>0\), which means \(w<12\). 3. Therefore the model makes sense for \(0<w<12\).

\(0<w<12\)

- Try counting a display with two rows of two pots. - Which pots are outside the square arrangement?

1. The display contains \(n\cdot n=n^2\) pots. 2. Include the four pots by the entrance: \(P(n)=n^2+4\).

\(P(n)=n^2+4\)

- Look for the highest point of the parabola. - Read its horizontal coordinate using the time axis. - Use the vertical axis label for the output unit.

1. The graph reaches its highest point at \((12,160)\). 2. The hour \(12\) is noon, and the y-coordinate represents power in watts. 3. The modeled maximum is \(160\,\text{W}\) at noon.

Noon; \(160\,\text{W}\)

- If each side were \(20\) feet long, what multiplication would give the area? - Replace that sample length with your chosen variable so the rule works for any side length. - Check that the result has square-foot units, not foot units.

1. Let \(s\) be the plaza's side length in feet. 2. Since each side has length \(s\), the area is \(A(s)=s^2\).

For example, let \(s\) be the side length in feet. Then \(A(s)=s^2\).

- Which variable represents time, and which represents distance? - How can you isolate the squared variable when the distance is given? - Compare both distances at the same time.

1. Evaluate Vehicle A's model: \(s_A(5) = 1.2 \cdot 5^2 = 1.2 \cdot 25 = 30\,\text{m}\). 2. Solve \(120 = 1.2t^2\): \(t^2 = 100\), so \(t = \pm 10\). Since time is nonnegative, \(t = 10\,\text{s}\). 3. Compare the distances at \(t = 10\): \(s_A(10) = 1.2 \cdot 10^2 = 120\,\text{m}\), while \(s_B(10) = 0.8 \cdot 10^2 + 2 \cdot 10 = 100\,\text{m}\). 4. Vehicle A is ahead because \(120\,\text{m} > 100\,\text{m}\).

a) \(30\,\text{m}\) b) \(10\,\text{s}\) c) Vehicle A; it has traveled \(120\,\text{m}\), compared with Vehicle B's \(100\,\text{m}\).

- What input represents the release point? - Which point of a downward-opening parabola gives the maximum value? - Substitute the specified horizontal distance into the model.

1. The release occurs at \(x = 0\): \(h(0) = 2\), so the release height is \(2\,\text{m}\). 2. The vertex occurs at \(x = -\frac{0.6}{2 \cdot (-0.05)} = 6\). Then \(h(6) = -0.05 \cdot 6^2 + 0.6 \cdot 6 + 2 = 3.8\). The maximum height is \(3.8\,\text{m}\). 3. Evaluate \(h(10) = -0.05 \cdot 10^2 + 0.6 \cdot 10 + 2 = 3\). The height is \(3\,\text{m}\).

a) \(2\,\text{m}\) b) \(3.8\,\text{m}\) c) \(3\,\text{m}\)

- Substitute each requested speed into the model. - What does equal output at two speeds suggest about the axis of symmetry? - Which formula gives the vertex of a quadratic in standard form?

1. Evaluate the model: \(g(40) = 0.002 \cdot 40^2 - 0.24 \cdot 40 + 10.7 = 4.3\), and \(g(80) = 0.002 \cdot 80^2 - 0.24 \cdot 80 + 10.7 = 4.3\). The two speeds have the same modeled fuel use. 2. Since the leading coefficient is positive, the minimum occurs at the vertex. Its speed is \(v = -\frac{-0.24}{2 \cdot 0.002} = 60\,\text{mph}\). 3. Evaluate \(g(60) = 0.002 \cdot 60^2 - 0.24 \cdot 60 + 10.7 = 3.5\). The minimum fuel use is \(3.5\) gallons per \(100\) miles.

a) \(4.3\) gallons per \(100\) miles at both \(40\,\text{mph}\) and \(80\,\text{mph}\) b) The minimum occurs at \(60\,\text{mph}\) and is \(3.5\) gallons per \(100\) miles.

- Substitute either measurement into \(s(v) = kv^2\). - Check the same value of \(k\) with the other measurement. - For part c, set the model equal to the given distance.

1. Use the first measurement: \(20 = k \cdot 20^2\), so \(k = \frac{20}{400} = 0.05\). 2. Check the second measurement: \(0.05 \cdot 40^2 = 80\), so both measurements fit the model \(s(v) = 0.05v^2\). 3. At \(60\,\text{mph}\), \(s(60) = 0.05 \cdot 60^2 = 180\). The predicted braking distance is \(180\,\text{ft}\). 4. Set \(0.05v^2 = 125\). Then \(v^2 = 2500\), so \(v = 50\) because speed is positive.

a) \(k = 0.05\), so \(s(v) = 0.05v^2\) b) \(180\,\text{ft}\) c) \(50\,\text{mph}\)

- Substitute each given horizontal distance into the function. - Where is the maximum of a downward-opening parabola? - What equation represents the points where the arch meets the ground?

1. Evaluate both heights: \(f(2) = -0.25 \cdot 2^2 + 2 \cdot 2 = 3\), and \(f(6) = -0.25 \cdot 6^2 + 2 \cdot 6 = 3\). Therefore, the arch is \(3\,\text{m}\) high at both locations. 2. The vertex occurs at \(x = -\frac{2}{2 \cdot (-0.25)} = 4\). Then \(f(4) = -0.25 \cdot 4^2 + 2 \cdot 4 = 4\), so the maximum height is \(4\,\text{m}\). 3. Set \(f(x) = 0\): \(-0.25x^2 + 2x = 0\). Factoring gives \(x(-0.25x + 2) = 0\), so the zeros are \(x = 0\) and \(x = 8\). The bases are \(8\,\text{m}\) apart.

a) \(f(2) = f(6) = 3\,\text{m}\) b) \(4\,\text{m}\) c) \(8\,\text{m}\), because the zeros are \(x = 0\) and \(x = 8\).

- Substitute each speed into the model. - Compare the factor by which the input changes with the factor by which the output changes. - If every output doubles, what happens to the coefficient?

1. Evaluate the model: \(s(25) = 0.04 \cdot 25^2 = 25\,\text{ft}\), and \(s(50) = 0.04 \cdot 50^2 = 100\,\text{ft}\). 2. The speed doubles from \(25\) to \(50\), while the braking distance changes from \(25\) to \(100\), which is multiplied by \(4\). This occurs because \((2v)^2 = 4v^2\). 3. Doubling every output doubles the coefficient: \(s_{\text{wet}}(v) = 2(0.04v^2) = 0.08v^2\).

a) \(25\,\text{ft}\) at \(25\,\text{mph}\); \(100\,\text{ft}\) at \(50\,\text{mph}\) b) Doubling the speed quadruples the braking distance. c) \(s_{\text{wet}}(v) = 0.08v^2\)

- Distinguish carefully between radius and diameter. - Substitute the known radius and mass into \(m(r) = kr^2\). - What scale factor for the radius has a square equal to \(4\)?

1. Substitute \(m = 1.2\) and \(r = 10\): \(1.2 = k \cdot 10^2 = 100k\). Therefore, \(k = 0.012\,\frac{\text{kg}}{\text{cm}^2}\). 2. A diameter of \(30\,\text{cm}\) gives a radius of \(15\,\text{cm}\). Then \(m(15) = 0.012 \cdot 15^2 = 2.7\,\text{kg}\). 3. Since mass is proportional to \(r^2\), multiplying the radius by \(2\) multiplies the mass by \(2^2 = 4\). The radius must be doubled.

a) \(k = 0.012\,\frac{\text{kg}}{\text{cm}^2}\), so \(m(r) = 0.012r^2\) b) \(2.7\,\text{kg}\) c) The radius must be doubled.

- For part a, locate where the curve meets the horizontal level \(8\). - Read the release height from the graph at \(t=0\). - On the descending side, identify when the curve returns to that release height and when it reaches the x-axis.

1. The graph reaches \(8\,\text{m}\) at \(t=1\) and \(t=3\). The curve is on or above that height between those times, so \(1\le t\le3\). 2. At release, the graph shows height \(5\,\text{m}\). 3. On the descending side, the graph returns to \(5\,\text{m}\) at \(t=4\) and reaches the ground at \(t=5\). 4. Therefore, after release and before ground contact, the ball is below its release height for \(4<t<5\).

a) \(1\le t\le3\) b) \(4<t<5\)

- Identify the vertex directly from the given form and connect both coordinates to the context. - Compare prices that are equally far from \(\$8\). What symmetry does the squared term create? - Compare a distance of \(1\) dollar from \(\$8\) with a distance of \(2\) dollars from \(\$8\). - Relate those squared distances to the outside coefficient.

1. The model is in vertex form, so its vertex is \((8,3200)\). 2. Because the leading coefficient is negative, the parabola opens downward. 3. Therefore, the model predicts maximum revenue of \(\$3200\) when the ticket price is \(\$8\). 4. At \(p=7\) or \(p=9\), the squared distance from \(8\) is \(1\), so revenue is \(\$50\) below the maximum: \(\$3150\). 5. At \(p=6\) or \(p=10\), the squared distance is \(4\), so revenue is \(4\cdot\$50=\$200\) below the maximum: \(\$3000\). 6. Thus, the coefficient \(-50\) means the modeled revenue decreases by \(\$50\) times the square of the dollar distance from the optimal price \(\$8\).

The model predicts maximum revenue of \(\$3200\) at a ticket price of \(\$8\). At \(\$7\) or \(\$9\), revenue is \(\$3150\); at \(\$6\) or \(\$10\), it is \(\$3000\). The coefficient \(-50\) scales the loss from the maximum by the square of the price's distance from \(\$8\).

- In factored form, each zero comes from one factor being zero. - Compare each zero with the stated contextual domain. - Translate the valid x-value from hundreds of units to units and interpret profit \(0\).

1. The factored form is zero when \(x=-2\) or \(x=12\). 2. The domain requires \(x\ge0\), so \(x=-2\) is not a possible production level. 3. The meaningful zero is \(x=12\), which represents \(1200\) units. 4. At that production level, modeled profit is \(0\), so it is a break-even point.

The mathematical zeros are \(x=-2\) and \(x=12\). Only \(x=12\) is meaningful, representing break-even production of \(1200\) units.

- Compare consecutive output changes before looking at the second differences. - For equally spaced inputs, connect a quadratic's constant second difference to its leading coefficient. - Use the row with \(x=0\) to identify one coefficient directly. - Substitute one additional table value to finish the model before predicting the next value.

1. The first differences are \(3\), \(7\), and \(11\), so the constant second difference is \(4\). 2. For unit x-steps in \(ax^2+bx+c\), the second difference is \(2a\). Thus, \(2a=4\), so \(a=2\). 3. Since \(C(0)=5\), \(c=5\). 4. Using \(C(1)=8\), \(2+b+5=8\), so \(b=1\). 5. Therefore, \(C(x)=2x^2+x+5\). 6. At \(x=4\), \(C(4)=2(4^2)+4+5=41\), which represents \(\$4100\).

a) \(a=2\) b) \(C(x)=2x^2+x+5\), and \(C(4)=41\), so the predicted cost is \(\$4100\).

- Because the measurements are noisy, do not force an exact quadratic through a selected three-point subset. - Use every ordered pair in the quadratic regression calculation. - Check whether the sign of the leading coefficient agrees with the overall rise-then-fall pattern in the data. - Keep the calculator's unrounded coefficients for the prediction, then round the final height as requested.

1. Enter all six ordered pairs into a quadratic regression calculation rather than forcing an exact quadratic through only three points. 2. The regression gives \(a\approx-0.5267857\), \(b\approx4.0939286\), and \(c\approx3.0107143\). 3. To the nearest thousandth, the model is \(y\approx-0.527x^2+4.094x+3.011\). 4. Using the unrounded regression coefficients at \(x=6\) gives \(y\approx8.61\). 5. Therefore, the predicted height is about \(8.6\,\text{cm}\).

\(y\approx-0.527x^2+4.094x+3.011\) Predicted height at \(x=6\): \(8.6\,\text{cm}\)

- For each requested input, evaluate the regression model before comparing it with the measurement. - Keep the subtraction order consistent with the stated definition of residual. - Use the sign to decide whether the observed value lies above or below the model prediction. - For the final comparison, use absolute residuals so positive and negative signs do not hide the size of an error.

1. At \(x=2\), the model predicts \(H(2)=-0.8(2^2)+6(2)+12=20.8\). The residual is \(20.1-20.8=-0.7\,\text{ft}\). 2. At \(x=4\), the model predicts \(H(4)=-0.8(4^2)+6(4)+12=23.2\). The residual is \(22.9-23.2=-0.3\,\text{ft}\). 3. A negative residual means the measured range is below the model prediction. 4. The other residuals are \(0.4\), \(0.5\), and \(-0.6\) feet at \(x=1\), \(x=3\), and \(x=5\), respectively. 5. The largest absolute residual is \(0.7\,\text{ft}\), at \(x=2\).

a) At \(x=2\): \(-0.7\,\text{ft}\); at \(x=4\): \(-0.3\,\text{ft}\) b) Both are negative, so both measured ranges are below the model's predictions. c) The measurement at \(x=2\), with absolute residual \(0.7\,\text{ft}\)

- If the width were \(6\) feet, how would you find the length? Now use a letter instead. - Write the other side length using the stated difference. - What two measurements are multiplied to get a rectangle's area?

1. Let \(w\) be the width in feet. Its length is \(w+5\) feet. 2. Area is width times length, so \(A(w)=w(w+5)\).

For example, let \(w\) be the width in feet. Then \(A(w)=w(w+5)\).

- A rectangle has two lengths and two widths in its perimeter. - Express the unknown length using the fencing total and \(w\). - Combine the two side lengths to model the area.

1. The perimeter condition is \(2w+2L=40\), where \(L\) is the length. 2. Solving gives \(L=20-w\). 3. The area is \(A(w)=wL=w(20-w)\).

a) \(20-w\) feet b) \(A(w)=w(20-w)\)

- Express the radius in terms of elapsed time. - Which area formula uses the radius of a circle? - Square the entire expression for the radius.

1. The radius after \(t\) minutes is \(r=2t\) feet. 2. Using circle area, \(A(t)=\pi r^2=\pi(2t)^2=4\pi t^2\), for \(0\le t\le3\).

\(A(t)=4\pi t^2\), for \(0\le t\le3\)

- What is the revenue from selling \(n\) cards at the given price? - Profit is the difference between revenue and cost. - Use parentheses so every cost term is subtracted.

1. Revenue from \(n\) sales is \(R(n)=6n\). 2. Profit equals revenue minus cost: \(P(n)=6n-(10+0.5n^2)\). 3. Simplifying gives \(P(n)=-0.5n^2+6n-10\).

\(P(n)=-0.5n^2+6n-10\)

- The profit is zero when one of the two factors is zero. - Separate the mathematical solutions from what can happen in the setting. - Compare each zero with the theater's capacity.

1. Setting \(P(n)=0\) gives \(n-20=0\) or \(n-100=0\). 2. The mathematical zeros are \(n=20\) and \(n=100\). 3. Since the theater can sell at most \(80\) tickets, only \(n=20\) is attainable.

Zeros: \(n=20\) and \(n=100\). Only \(20\) tickets is a possible break-even sales level.

- Can you write the height using only the base length? - How is a triangle's area related to a rectangle with the same base and height? - Check that your rule gives an area, measured in square feet.

1. Let \(b\) be the base length in feet. The height is \(2b+3\) feet. 2. The area of a triangle is half the base times the height. 3. Therefore, \(A(b)=\frac{1}{2}b(2b+3)\).

For example, let \(b\) be the base in feet. Then \(A(b)=\frac{1}{2}b(2b+3)\).

- Which price is charged for covering a surface, and which for its boundary? - If one side doubled, how would each of those measurements change? - Work out the two charges separately, then combine them.

1. Let \(s\) be the side length in feet. 2. The area is \(s^2\) square feet, so the paving stones cost \(4s^2\) dollars. 3. The perimeter is \(4s\) feet, so the edging costs \(3(4s)=12s\) dollars. 4. The total material cost is \(C(s)=4s^2+12s\).

For example, let \(s\) be the side length in feet. Then \(C(s)=4s^2+12s\) dollars.

- Model each game as a pair of different players. - Make sure each matchup is counted only once. - Write the number of pairs in terms of \(n\). - Interpret both algebraic solutions in the context.

1. Let \(n\) be the number of players. Each game corresponds to an unordered pair of players, so the total number of games is \(\frac{n(n-1)}{2}\). 2. Set up the equation: \(\frac{n(n-1)}{2}=190\). 3. Rearrange: \(n^2-n-380=0\). 4. Apply the quadratic formula: \(n=\frac{1\pm\sqrt{(-1)^2-4\cdot1\cdot(-380)}}{2}=\frac{1\pm39}{2}\). 5. The solutions are \(n=20\) and \(n=-19\). Only the positive value is meaningful, so there are \(20\) players.

\(20\) players

- What does \(n\) represent in this model? - When an output value is given, what equation should you solve? - What restrictions does the context place on \(n\)?

1. Evaluate the model at \(n = 8\): \(d(8) = \frac{1}{2} \cdot 8^2 - 1.5 \cdot 8 = 32 - 12 = 20\). 2. Set the model equal to \(54\): \(54 = \frac{1}{2}n^2 - 1.5n\). Multiplying by \(2\) gives \(n^2 - 3n - 108 = 0\). 3. Factor: \((n - 12)(n + 9) = 0\), so \(n = 12\) or \(n = -9\). 4. A polygon cannot have a negative number of sides, so only \(n = 12\) is meaningful.

a) \(20\) diagonals b) \(12\) sides c) The solution \(n = -9\) is not meaningful because a polygon cannot have a negative number of sides.

- How many opponents does one team have? - Why does counting games team by team count each game twice? - What games must the new team play against the existing teams?

1. Each team has \(n - 1\) opponents and plays each opponent twice, giving \(2(n - 1)\) team-game counts per team. Multiplying by \(n\) counts each actual game twice, so \(S(n) = \frac{n \cdot 2(n - 1)}{2} = n(n - 1) = n^2 - n\). 2. Solve \(n^2 - n = 240\): \(n^2 - n - 240 = 0\). Factoring gives \((n - 16)(n + 15) = 0\), so \(n = 16\) or \(n = -15\). Only \(n = 16\) is meaningful. 3. The increase is \(S(n + 1) - S(n) = [(n + 1)^2 - (n + 1)] - (n^2 - n) = 2n\). Equivalently, the new team plays two games against each of the \(n\) existing teams.

a) \(S(n) = n(n - 1) = n^2 - n\) b) \(16\) teams c) \(2n\) additional games

- Use the formula for the number of pairs in a group. - Solve the equation in part a) before doubling the group size. - Recalculate the model for the doubled number of computers. - Compare the new result with twice the original value.

1. If there are \(n\) computers, the number of cables is \(\frac{n(n-1)}{2}\). For part a), solve \(\frac{n(n-1)}{2}=66\). 2. Rearranging gives \(n^2-n-132=0\). The quadratic formula gives \(n=\frac{1\pm23}{2}\), so \(n=12\) or \(n=-11\). Thus, the network has \(12\) computers. 3. Doubling the number of computers gives \(24\) computers. The required number of cables is \(\frac{24\cdot23}{2}=276\). 4. Twice the original number of cables is \(2\cdot66=132\). Since \(276\ne132\), the claim is false. The number of connections follows a quadratic, not linear, model.

a) \(12\) computers b) The claim is false. With \(24\) computers, the network needs \(276\) cables, not \(132\).

- Consecutive odd integers differ by \(2\). - Express both integers using one variable before using the product condition. - A positive product can come from two positive numbers or two negative numbers.

1. If \(x\) is the smaller odd integer, the next odd integer is \(x+2\). 2. The product condition gives \(x(x+2)=195\), so \(x^2+2x-195=0\). 3. Factor: \((x-13)(x+15)=0\), so \(x=13\) or \(x=-15\). 4. The corresponding ordered pairs are \((13,15)\) and \((-15,-13)\).

a) \(x^2+2x-195=0\) b) \((-15,-13)\) and \((13,15)\)

- Express the length in terms of the width. - Use the area condition to create one equation in one variable. - Check which algebraic solution can represent a physical width. - The fencing length is the perimeter of the final rectangle.

1. Let \(x\) be the width in meters. Then the length is \(x+10\). 2. The area condition is \(x(x+10)=375\), so \(x^2+10x-375=0\). 3. Factor: \((x-15)(x+25)=0\), so \(x=15\) or \(x=-25\). 4. A width must be positive, so the lot is \(15\,\text{m}\) by \(25\,\text{m}\). 5. The perimeter is \(2(15+25)=80\,\text{m}\).

a) \(x^2+10x-375=0\) b) \(15\,\text{m}\) by \(25\,\text{m}\) c) \(80\,\text{m}\)

- Express the shorter leg in terms of the longer leg. - The two leg expressions and the hypotenuse must satisfy one relationship. - After solving, reject any value that cannot be a side length. - Use the two valid legs as the base and height for the area.

1. Let \(x\) be the longer leg in centimeters. The shorter leg is \(x-5\). 2. The Pythagorean theorem gives \(x^2+(x-5)^2=25^2\). 3. Expand and simplify: \(2x^2-10x-600=0\), so \(x^2-5x-300=0\). 4. Factor: \((x-20)(x+15)=0\), so \(x=20\) or \(x=-15\). 5. The valid legs are \(20\,\text{cm}\) and \(15\,\text{cm}\). 6. The area is \(\frac{1}{2}\cdot20\cdot15=150\,\text{cm}^2\).

a) \(x^2+(x-5)^2=25^2\), equivalently \(x^2-5x-300=0\) b) \(20\,\text{cm}\) and \(15\,\text{cm}\) c) \(150\,\text{cm}^2\)

- What input represents the nozzle location? - How do you find the vertex of a quadratic model? - What value must \(h(x)\) have when the water reaches the surface? - Which root is meaningful as a distance from the nozzle?

1. At the nozzle, \(x = 0\): \(h(0) = 1.25\). The nozzle is \(1.25\,\text{m}\) above the water surface. 2. The vertex occurs at \(x = -\frac{1}{2 \cdot (-0.25)} = 2\). Then \(h(2) = -0.25 \cdot 2^2 + 2 + 1.25 = 2.25\). The highest point is \((2, 2.25)\). 3. Set \(h(x) = 0\): \(-0.25x^2 + x + 1.25 = 0\). Multiplying by \(-4\) gives \(x^2 - 4x - 5 = 0\), or \((x - 5)(x + 1) = 0\). 4. The solutions are \(x = 5\) and \(x = -1\). Only the positive distance is meaningful, so the water hits the surface \(5\,\text{m}\) from the nozzle.

a) \(1.25\,\text{m}\) b) \((2, 2.25)\) c) \(5\,\text{m}\)

- Represent the second number in terms of the smaller number. - Translate the sum-of-squares condition into one equation before simplifying. - After solving the quadratic, check which root satisfies the natural-number condition.

1. The next natural number is \(n+1\), so the condition gives \(n^2+(n+1)^2=313\). 2. Expand and simplify: \(2n^2+2n-312=0\), so \(n^2+n-156=0\). 3. Solve the quadratic: \(n=\frac{-1\pm\sqrt{1-4(1)(-156)}}{2}=\frac{-1\pm25}{2}\). 4. Thus, \(n=12\) or \(n=-13\). Only \(12\) is a natural number, so the two numbers are \(12\) and \(13\).

The quadratic equation is \(n^2+n-156=0\), and the two natural numbers are \(12\) and \(13\).

- Express both new rectangle dimensions in terms of the original square side. - Use the new area to create one equation in \(x\). - Check which algebraic solution can represent a side length. - Compare the original square area with the new rectangle area.

1. Let \(x\) be the side length of the square. The rectangle has side lengths \(x+3\) and \(x-2\). 2. The area condition is \((x+3)(x-2)=84\). 3. Expanding and rearranging gives \(x^2+x-90=0\). 4. Factor: \((x-9)(x+10)=0\), so \(x=9\) or \(x=-10\). Only \(x=9\,\text{cm}\) is a valid length. 5. The original area was \(9^2=81\,\text{cm}^2\), so the increase was \(84-81=3\,\text{cm}^2\).

a) \((x+3)(x-2)=84\), equivalently \(x^2+x-90=0\) b) \(9\,\text{cm}\) c) \(3\,\text{cm}^2\)

- Write each square area in terms of its side length. - Express the larger side using the smaller side. - Your quadratic has two mathematical roots; connect each root back to the side-length variable. - Which root satisfies the physical constraint on a length?

1. Let \(s\) be the smaller side length in centimeters. The larger side length is \(s+10\). 2. The sum of the areas gives \(s^2+(s+10)^2=2500\). 3. Simplify: \(2s^2+20s-2400=0\), or \(s^2+10s-1200=0\). 4. Factor: \((s-30)(s+40)=0\), so \(s=30\) or \(s=-40\). 5. A side length cannot be negative, so the smaller tile has side \(30\,\text{cm}\), and the larger tile has side \(40\,\text{cm}\).

a) \(s^2+(s+10)^2=2500\), equivalently \(s^2+10s-1200=0\) b) The roots are \(s=30\) and \(s=-40\), so the tile side lengths are \(30\,\text{cm}\) and \(40\,\text{cm}\). c) The root \(s=-40\) is rejected because a physical length cannot be negative.

- What is the value of \(h_0\) at launch? - Substitute the known time and height to solve for the missing coefficient. - Once the model is complete, substitute the requested time.

1. The initial height is \(h_0 = 1.5\). Substitute \(t = 2\) and \(h(2) = 31.5\): \(31.5 = -5 \cdot 2^2 + 2v_0 + 1.5\). 2. Simplify: \(31.5 = -18.5 + 2v_0\), so \(50 = 2v_0\) and \(v_0 = 25\). 3. The height function is \(h(t) = -5t^2 + 25t + 1.5\). 4. Evaluate at \(t = 4\): \(h(4) = -5 \cdot 4^2 + 25 \cdot 4 + 1.5 = 21.5\).

a) \(h(t) = -5t^2 + 25t + 1.5\) b) \(21.5\,\text{m}\)

- What equation follows from being at the same height at \(t = 0\) and \(t = 0.8\)? - How does parabola symmetry locate the vertex between two equal-height times? - Evaluate the completed model at that time.

1. Since \(h_0 = 10\) and \(h(0.8) = 10\), substitute into the model: \(10 = -5 \cdot 0.8^2 + 0.8v_0 + 10\). 2. Simplify: \(0 = -3.2 + 0.8v_0\), so \(v_0 = 4\). Thus, \(h(t) = -5t^2 + 4t + 10\). 3. The equal heights at \(t = 0\) and \(t = 0.8\) are symmetric about the vertex, so the maximum occurs at \(t = 0.4\,\text{s}\). 4. Evaluate \(h(0.4) = -5 \cdot 0.4^2 + 4 \cdot 0.4 + 10 = 10.8\). The maximum height is \(10.8\,\text{m}\).

The maximum height is \(10.8\,\text{m}\).

- First find the maximum of the fully specified model. - How does the vertex-time formula determine Ball B's initial-velocity coefficient? - Evaluate both models at their vertex times and compare.

1. Ball A reaches its vertex at \(t = -\frac{20}{2 \cdot (-5)} = 2\). Its maximum height is \(h_A(2) = -5 \cdot 2^2 + 20 \cdot 2 + 2 = 22\,\text{m}\). 2. For Ball B, the vertex time satisfies \(1.5 = -\frac{v_0}{2 \cdot (-5)} = \frac{v_0}{10}\), so \(v_0 = 15\). 3. Ball B's model is \(h_B(t) = -5t^2 + 15t + 1\). Its maximum height is \(h_B(1.5) = -5 \cdot 1.5^2 + 15 \cdot 1.5 + 1 = 12.25\,\text{m}\). 4. Since \(22\,\text{m} > 12.25\,\text{m}\), Ball A reaches the greater maximum height.

Ball A reaches the greater maximum height: \(22\,\text{m}\), compared with Ball B's \(12.25\,\text{m}\).

- What input represents the release point? - Where is the maximum of a downward-opening quadratic? - Evaluate the model at the obstacle's horizontal position. - What comparison determines whether the ball clears the obstacle?

1. At release, \(x = 0\): \(h(0) = 2\), so the release height is \(2\,\text{m}\). 2. The vertex occurs at \(x = -\frac{1}{2 \cdot (-0.1)} = 5\). Then \(h(5) = -0.1 \cdot 5^2 + 5 + 2 = 4.5\). The maximum height is \(4.5\,\text{m}\). 3. Evaluate the model at \(x = 9\): \(h(9) = -0.1 \cdot 9^2 + 9 + 2 = 2.9\). Since \(2.9\,\text{m} < 3\,\text{m}\), the ball does not clear the obstacle.

a) \(2\,\text{m}\) b) \(4.5\,\text{m}\) c) No. At \(x = 9\), the ball is \(2.9\,\text{m}\) high.

- How do you find the maximum of each downward-opening quadratic? - What does a positive zero represent in this context? - Keep only the positive solution because distance is measured forward from \(x = 0\). - Compare the two landing distances.

1. Athlete A's vertex occurs at \(x = -\frac{0.6}{2 \cdot (-0.05)} = 6\). The maximum height is \(h_A(6) = 3.8\,\text{m}\). 2. Athlete B's vertex occurs at \(x = -\frac{0.5}{2 \cdot (-0.04)} = 6.25\). The maximum height is \(h_B(6.25) = 3.6625\,\text{m}\). Therefore, Athlete A's shot reaches the greater maximum height. 3. For Athlete A, solve \(-0.05x^2 + 0.6x + 2 = 0\). The positive zero is \(x = 6 + \sqrt{76} \approx 14.72\,\text{m}\). 4. For Athlete B, solve \(-0.04x^2 + 0.5x + 2.1 = 0\). The positive zero is \(x = \frac{25 + \sqrt{1465}}{4} \approx 15.82\,\text{m}\). 5. Since \(15.82\,\text{m} > 14.72\,\text{m}\), Athlete B's shot travels farther.

a) Athlete A, with a maximum height of \(3.8\,\text{m}\), compared with Athlete B's \(3.6625\,\text{m}\) b) Athlete B; its shot travels about \(15.82\,\text{m}\), compared with Athlete A's \(14.72\,\text{m}\).

- Draw the enclosure and identify which three sides require fencing. - Use the fencing constraint to express one dimension in terms of the other. - Write the area as a quadratic function and find its vertex. - For the square comparison, all three fenced sides have the same length.

1. Let \(x\) be the length of each side perpendicular to the barn and \(y\) the fenced side parallel to the barn. The fencing constraint is \(2x + y = 120\), so \(y = 120 - 2x\). 2. The area is \(A(x) = x(120 - 2x) = -2x^2 + 120x\). 3. The maximum occurs at the vertex: \(x = -\frac{120}{2 \cdot (-2)} = 30\). Then \(y = 120 - 2 \cdot 30 = 60\). 4. The maximum area is \(30 \cdot 60 = 1800\,\text{ft}^2\). 5. For a square with three fenced sides, \(3s = 120\), so \(s = 40\,\text{ft}\). Its area is \(40^2 = 1600\,\text{ft}^2\), which is smaller than the maximum rectangular area.

The maximizing dimensions are \(30\,\text{ft}\) perpendicular to the barn and \(60\,\text{ft}\) parallel to it, giving \(1800\,\text{ft}^2\). A square would have side length \(40\,\text{ft}\) and area \(1600\,\text{ft}^2\), so it has less area.

- Let an integer variable count the number of 10-cent price changes. - Revenue equals price times quantity sold. - The vertex lies between two allowable integer inputs, so compare both neighboring values.

1. Let \(x\) be the integer number of \(\$0.10\) price increases, where a negative value represents a decrease. The price is \(P(x) = 1.50 + 0.10x\), and the number sold is \(Q(x) = 200 - 10x\). 2. The revenue is \(R(x) = (1.50 + 0.10x)(200 - 10x) = -x^2 + 5x + 300\). 3. The continuous vertex occurs at \(x = -\frac{5}{2 \cdot (-1)} = 2.5\). Because \(x\) must be an integer, compare \(x = 2\) and \(x = 3\). 4. \(R(2) = 306\) and \(R(3) = 306\). The corresponding prices are \(P(2) = \$1.70\) and \(P(3) = \$1.80\). Both maximize the daily revenue.

The maximizing prices are \(\$1.70\) and \(\$1.80\). Each gives daily revenue of \(\$306\).

- Profit equals revenue minus fixed cost. - Use the vertex of the profit function for part a. - For part b, find the zeros and determine where the downward-opening parabola is on or above the x-axis.

1. Revenue is \(xN(x)\), so profit is \(P(x) = x(200 - 10x) - 750 = -10x^2 + 200x - 750\). 2. The maximum occurs at the vertex: \(x = -\frac{200}{2 \cdot (-10)} = 10\). The profit-maximizing ticket price is \(\$10\). 3. To find the no-loss interval, solve \(-10x^2 + 200x - 750 = 0\). Dividing by \(-10\) gives \(x^2 - 20x + 75 = 0\). 4. Factor: \((x - 5)(x - 15) = 0\). The break-even prices are \(\$5\) and \(\$15\). 5. Since the profit parabola opens downward, profit is nonnegative between the zeros, including the endpoints. Thus, \(5 \leq x \leq 15\).

a) \(\$10\) b) Ticket prices from \(\$5\) through \(\$15\), including both break-even prices

- Evaluate both models at the same speed. - Use the dry-pavement distance as the original amount when finding the percent increase. - For part b, first find the wet-pavement distance, then set the dry model equal to it.

1. At \(30\,\text{mph}\), \(s_d(30) = 0.04 \cdot 30^2 = 36\,\text{ft}\), and \(s_w(30) = 0.064 \cdot 30^2 = 57.6\,\text{ft}\). 2. The increase is \(57.6 - 36 = 21.6\,\text{ft}\). Relative to the dry distance, the percent increase is \(\frac{21.6}{36} \cdot 100\% = 60\%\). 3. The wet-pavement distance at \(50\,\text{mph}\) is \(s_w(50) = 0.064 \cdot 50^2 = 160\,\text{ft}\). 4. Set the dry-pavement model equal to \(160\): \(0.04v^2 = 160\). Then \(v^2 = 4000\), so \(v = \sqrt{4000} \approx 63.2\).

a) Dry: \(36\,\text{ft}\); wet: \(57.6\,\text{ft}\). The wet-pavement distance is \(60\%\) greater. b) Approximately \(63.2\,\text{mph}\)

- Evaluate each term of the model separately before adding. - Replace \(v\) with \(2v\) to compare the two parts of the model. - In part c, use only the quadratic braking-distance term.

1. Evaluate the model at \(65\): \(S(65) = 1.1 \cdot 65 + 0.04 \cdot 65^2 = 71.5 + 169 = 240.5\). The total stopping distance is \(240.5\,\text{ft}\). 2. If the speed changes from \(v\) to \(2v\), the reaction term becomes \(1.1(2v) = 2(1.1v)\), so it doubles. The braking term becomes \(0.04(2v)^2 = 4(0.04v^2)\), so it quadruples. 3. Set the braking term equal to \(144\): \(0.04v^2 = 144\). Then \(v^2 = 3600\), so \(v = 60\) because speed is positive.

a) \(240.5\,\text{ft}\) b) Doubling the speed doubles the reaction-distance term and quadruples the braking-distance term. c) \(60\,\text{mph}\)

- Where is the maximum of a downward-opening parabola? - What equation represents the moment the ball reaches the ground? - Which coefficient gives the height when \(x = 0\)?

1. The maximum occurs at the vertex. Its horizontal coordinate is \(x = -\frac{2}{2 \cdot (-0.5)} = 2\). 2. Evaluate the height: \(h(2) = -0.5 \cdot 2^2 + 2 \cdot 2 + 2.5 = 4.5\). The maximum height is \(4.5\,\text{m}\). 3. To find when the ball reaches the ground, solve \(-0.5x^2 + 2x + 2.5 = 0\). Multiplying by \(-2\) gives \(x^2 - 4x - 5 = 0\), which factors as \((x - 5)(x + 1) = 0\). The context-valid solution is \(x = 5\), so the horizontal distance is \(5\,\text{m}\). 4. Raising the initial height by \(1\,\text{m}\) increases the constant term from \(2.5\) to \(3.5\). The new function is \(h_{\text{new}}(x) = -0.5x^2 + 2x + 3.5\).

a) \(4.5\,\text{m}\) b) \(5\,\text{m}\) c) \(h_{\text{new}}(x) = -0.5x^2 + 2x + 3.5\)

- How many points determine a linear model? How many determine a quadratic model? - Substitute the data points into each model's general form. - Compare each week-\(5\) prediction with the measured height using absolute error.

1. For the linear model, \(h_{\text{lin}}(0) = 50\), so the y-intercept is \(50\). The slope through \((0, 50)\) and \((4, 170)\) is \(\frac{170 - 50}{4 - 0} = 30\). Thus, \(h_{\text{lin}}(t) = 30t + 50\). 2. Let \(h_{\text{quad}}(t) = at^2 + bt + c\). From \(h_{\text{quad}}(0) = 50\), \(c = 50\). 3. Using \((2, 90)\) gives \(4a + 2b = 40\), or \(2a + b = 20\). Using \((4, 170)\) gives \(16a + 4b = 120\), or \(4a + b = 30\). 4. Subtract the equations: \(2a = 10\), so \(a = 5\). Then \(2 \cdot 5 + b = 20\), so \(b = 10\). Therefore, \(h_{\text{quad}}(t) = 5t^2 + 10t + 50\). 5. At week \(5\), the linear model predicts \(h_{\text{lin}}(5) = 200\,\text{cm}\), with an absolute error of \(22.5\,\text{cm}\). The quadratic model predicts \(h_{\text{quad}}(5) = 225\,\text{cm}\), with an absolute error of \(2.5\,\text{cm}\). The quadratic model gives the better prediction.

a) \(h_{\text{lin}}(t) = 30t + 50\) \(h_{\text{quad}}(t) = 5t^2 + 10t + 50\) b) The quadratic model is better. Its prediction is \(225\,\text{cm}\), which is \(2.5\,\text{cm}\) from the measured value; the linear prediction is \(200\,\text{cm}\), which is \(22.5\,\text{cm}\) away.

- Use the point \((10, 80)\) to find the slope of the linear model. - Substitute the two nonzero data points into \(g(x) = ax^2 + bx\). - Compare whether each model has a constant or increasing rate of change.

1. For the linear model, substitute \((10, 80)\): \(80 = 10m\), so \(m = 8\). Therefore, \(f(x) = 8x\). 2. For the quadratic model, use \((5, 25)\): \(25a + 5b = 25\), so \(5a + b = 5\). 3. Use \((10, 80)\): \(100a + 10b = 80\), so \(10a + b = 8\). Subtracting the equations gives \(5a = 3\), so \(a = 0.6\). Then \(5 \cdot 0.6 + b = 5\), so \(b = 2\). Thus, \(g(x) = 0.6x^2 + 2x\). 4. In month \(12\), \(f(12) = 8 \cdot 12 = 96\), so the linear model predicts \(96{,}000\) users. Also, \(g(12) = 0.6 \cdot 12^2 + 2 \cdot 12 = 110.4\), so the quadratic model predicts \(110{,}400\) users. 5. A linear model has a constant growth rate, while this quadratic model has an increasing growth rate for \(x \geq 0\). Under the stated assumption, the quadratic model is more consistent.

a) \(f(x) = 8x\) b) \(g(x) = 0.6x^2 + 2x\) c) Linear model: \(96{,}000\) users Quadratic model: \(110{,}400\) users The quadratic model is more consistent with an increasing growth rate.

- Evaluate both functions at \(x = 0\). - Find the vertex of each downward-opening parabola. - To find horizontal distance, solve each equation when the height is \(0\) and keep the positive solution.

1. At release, \(x = 0\). Since \(h_1(0) = h_2(0) = 6\), both balls are released from \(6\,\text{ft}\). 2. For the first ball, the vertex occurs at \(x = -\frac{1.2}{2 \cdot (-0.1)} = 6\). Its maximum height is \(h_1(6) = -0.1 \cdot 6^2 + 1.2 \cdot 6 + 6 = 9.6\,\text{ft}\). 3. For the second ball, the vertex occurs at \(x = -\frac{0.8}{2 \cdot (-0.05)} = 8\). Its maximum height is \(h_2(8) = -0.05 \cdot 8^2 + 0.8 \cdot 8 + 6 = 9.2\,\text{ft}\). Therefore, the first ball reaches the greater maximum height. 4. For the first ball, solve \(-0.1x^2 + 1.2x + 6 = 0\). This is equivalent to \(x^2 - 12x - 60 = 0\), so the positive zero is \(x = 6 + 4\sqrt{6} \approx 15.80\,\text{ft}\). 5. For the second ball, solve \(-0.05x^2 + 0.8x + 6 = 0\). This is equivalent to \(x^2 - 16x - 120 = 0\), so the positive zero is \(x = 8 + 2\sqrt{46} \approx 21.56\,\text{ft}\). Therefore, the second ball travels farther.

a) \(6\,\text{ft}\) b) The first ball; its maximum height is \(9.6\,\text{ft}\), compared with \(9.2\,\text{ft}\) for the second ball. c) The second ball travels farther. The first travels about \(15.80\,\text{ft}\), and the second travels about \(21.56\,\text{ft}\).

- What profit value represents breaking even? - Which feature of a downward-opening parabola gives the maximum profit? - Interpret the input and output when \(x=0\).

1. At a break-even price, \(P(x)=0\). Solve \(-2x^2+60x-400=0\). Dividing by \(-2\) gives \(x^2-30x+200=0\), which factors as \((x-10)(x-20)=0\). The break-even prices are \(\$10\) and \(\$20\). 2. The maximum occurs at the vertex because the parabola opens downward. Its horizontal coordinate is \(x=-\frac{60}{2\cdot(-2)}=15\). Then \(P(15)=-2\cdot15^2+60\cdot15-400=50\). The profit is maximized at a price of \(\$15\), and the maximum modeled monthly profit is \(\$50\). 3. The value \(P(0)=-400\) means that the model predicts a \(\$400\) monthly loss if the shirts are priced at \(\$0\). The model alone does not show that the entire amount represents fixed costs.

a) \(\$10\) and \(\$20\) b) A price of \(\$15\) gives a maximum monthly profit of \(\$50\). c) The model predicts a monthly loss of \(\$400\) when the price is \(\$0\); the model does not establish that this amount consists entirely of fixed costs.

- Use the given sum to express one leg in terms of the other. - Apply the Pythagorean theorem. - Expand the square of a difference carefully. - Factor the resulting quadratic equation.

1. Let one leg be \(a\). Then the other leg is \(31-a\). 2. Apply the Pythagorean theorem: \(a^2+(31-a)^2=25^2\). 3. Expand and simplify: \(2a^2-62a+336=0\), so \(a^2-31a+168=0\). 4. Factor: \((a-7)(a-24)=0\), so \(a=7\) or \(a=24\). 5. The corresponding other leg is \(24\) or \(7\). Therefore, the legs are \(7\,\text{cm}\) and \(24\,\text{cm}\).

\(7\,\text{cm}\) and \(24\,\text{cm}\)

- Express the longer leg in terms of the shorter leg. - Use the two leg expressions in the triangle area formula. - Check which algebraic solution can represent a length. - Once the legs are known, use their right-angle relationship to find the remaining side.

1. Let the shorter leg be \(x\). The longer leg is \(x+7\). 2. The area condition is \(30=\frac{1}{2}x(x+7)\), so \(x^2+7x-60=0\). 3. Factor: \((x-5)(x+12)=0\), so \(x=5\) or \(x=-12\). 4. A leg length must be positive, so the legs are \(5\,\text{cm}\) and \(12\,\text{cm}\). 5. By the Pythagorean theorem, \(c^2=5^2+12^2=169\), so \(c=13\,\text{cm}\).

a) \(x^2+7x-60=0\) b) \(5\,\text{cm}\) and \(12\,\text{cm}\) c) \(13\,\text{cm}\)

- What value of \(y\) represents ground level? - For part b, what height corresponds to the vehicle's top corners? - Because the vehicle is centered, its width spans equal distances on both sides of the y-axis.

1. At ground level, \(y = 0\). Solve \(0 = -0.2x^2 + 5\): \(0.2x^2 = 5\), so \(x^2 = 25\) and \(x = \pm 5\). The tunnel is \(5 - (-5) = 10\,\text{m}\) wide at the ground. 2. At the top corners of a \(4\,\text{m}\)-high vehicle, set \(y = 4\): \(4 = -0.2x^2 + 5\). Then \(0.2x^2 = 1\), so \(x^2 = 5\) and \(x = \pm\sqrt{5}\). 3. The greatest possible width is the distance between these two horizontal positions: \(\sqrt{5} - (-\sqrt{5}) = 2\sqrt{5} \approx 4.47\,\text{m}\).

a) \(10\,\text{m}\) b) \(2\sqrt{5}\,\text{m} \approx 4.47\,\text{m}\)

- Split the total distance equally on both sides of the y-axis. - Substitute a station's horizontal coordinate into the model. - At the water elevation, find both symmetric horizontal coordinates and calculate the full width.

1. Because the stations are at equal heights on opposite sides of the y-axis and are \(120\,\text{m}\) apart, their horizontal coordinates are \(x = -60\) and \(x = 60\). 2. Evaluate the model: \(h = 0.005 \cdot 60^2 = 18\). The stations are \(18\,\text{m}\) above the lowest point. 3. The water elevation is \(\frac{1}{2} \cdot 18 = 9\,\text{m}\). Set \(9 = 0.005x^2\), so \(x^2 = 1800\) and \(x = \pm\sqrt{1800} = \pm 30\sqrt{2}\). 4. The water surface extends from \(-30\sqrt{2}\) to \(30\sqrt{2}\), so its width is \(60\sqrt{2}\,\text{m} \approx 84.85\,\text{m}\).

a) \(18\,\text{m}\) b) \(60\sqrt{2}\,\text{m} \approx 84.85\,\text{m}\)

- A uniform frame adds the width on both sides of each photo dimension. - Use the outer rectangle's area to form one equation in \(x\). - A quadratic model can produce a mathematical root that violates the physical meaning of the variable.

1. The outer dimensions are \(20+2x\) and \(30+2x\). 2. The area condition is \((20+2x)(30+2x)=1064\). 3. Expanding and simplifying gives \(4x^2+100x-464=0\), or \(x^2+25x-116=0\). 4. The quadratic formula gives \(x=\frac{-25\pm33}{2}\), so \(x=4\) or \(x=-29\). 5. A frame width cannot be negative, so the frame width is \(4\,\text{cm}\).

a) \((20+2x)(30+2x)=1064\), equivalently \(x^2+25x-116=0\) b) \(4\,\text{cm}\) c) The root \(x=-29\) is rejected because a frame width cannot be negative.

- Use all \(60\,\text{m}\) of fencing to relate the two side lengths. - Substitute that relationship into the area condition. - After solving, pair each root with the corresponding value of the other side. - Compare the two resulting ordered side pairs.

1. If one side is \(x\), the perimeter condition gives \(x+y=30\), so the other side is \(30-x\). 2. The area condition is \(x(30-x)=221\), so \(x^2-30x+221=0\). 3. Factor: \((x-13)(x-17)=0\), so \(x=13\) or \(x=17\). 4. When \(x=13\), the other side is \(17\); when \(x=17\), the other side is \(13\). 5. Both roots therefore describe the same \(13\,\text{m}\) by \(17\,\text{m}\) rectangle with the side labels exchanged.

a) The other side is \(30-x\), and \(x(30-x)=221\), equivalently \(x^2-30x+221=0\). b) \(13\,\text{m}\) by \(17\,\text{m}\) c) The roots exchange which side is called \(x\), so they describe the same rectangle.

- An inside border reduces each center dimension on two sides. - Before solving, ask how wide the border can be while leaving a positive center rectangle. - Compare every algebraic solution with that physical domain.

1. The center rectangle has dimensions \(15-2x\) and \(10-2x\). 2. The area condition is \((15-2x)(10-2x)=84\), which simplifies to \(2x^2-25x+33=0\). 3. Because the smaller outside dimension is \(10\,\text{ft}\), the center must have positive width, so \(0<x<5\). 4. The quadratic formula gives \(x=\frac{25\pm19}{4}\), so \(x=11\) or \(x=1.5\). 5. Only \(x=1.5\) lies in the physical domain. The border width is \(1.5\,\text{ft}\).

a) \((15-2x)(10-2x)=84\), equivalently \(2x^2-25x+33=0\), with \(0<x<5\) b) \(1.5\,\text{ft}\)

- Use the two ground-level points as the zeros of the quadratic. - Substitute the additional point to determine the leading coefficient. - Where is the vertex relative to two symmetric zeros?

1. The arch has zeros at \(x = 0\) and \(x = 40\), so write \(h(x) = ax(x - 40)\). 2. Use the point \((10, 15)\): \(15 = a \cdot 10 \cdot (10 - 40) = -300a\). Thus, \(a = -0.05\). 3. The model is \(h(x) = -0.05x(x - 40) = -0.05x^2 + 2x\). 4. The vertex lies midway between the zeros at \(x = 20\). Evaluate \(h(20) = -0.05 \cdot 20^2 + 2 \cdot 20 = 20\). The maximum height is \(20\,\text{m}\).

a) \(h(x) = -0.05x^2 + 2x\) b) \(20\,\text{m}\)

- Use the starting and landing positions as zeros. - The maximum lies midway between the two zeros. - For part c, set the function equal to the specified height and solve the resulting quadratic equation.

1. The zeros are \(x = 0\) and \(x = 8\), so write \(f(x) = ax(x - 8)\). 2. Use the point \((2, 3)\): \(3 = a \cdot 2 \cdot (2 - 8) = -12a\), so \(a = -0.25\). Thus, \(f(x) = -0.25x^2 + 2x\). 3. The vertex lies midway between the zeros at \(x = 4\). Evaluate \(f(4) = -0.25 \cdot 4^2 + 2 \cdot 4 = 4\). The maximum height is \(4\,\text{m}\). 4. Set the height equal to \(2\): \(-0.25x^2 + 2x = 2\). This is equivalent to \(x^2 - 8x + 8 = 0\). 5. Using the quadratic formula, \(x = \frac{8 \pm \sqrt{64 - 32}}{2} = 4 \pm 2\sqrt{2}\). Therefore, \(x \approx 1.17\,\text{m}\) and \(x \approx 6.83\,\text{m}\).

a) \(f(x) = -0.25x^2 + 2x\) b) \(4\,\text{m}\) c) Approximately \(1.17\,\text{m}\) and \(6.83\,\text{m}\)

- Use the perimeter to relate the two side lengths. - Express one side in terms of the other and write the area as a quadratic. - The vertex of the downward-opening area function gives the maximum.

1. The \(8\)-by-\(12\) bed has area \(8 \cdot 12 = 96\,\text{ft}^2\). 2. A square with perimeter \(40\,\text{ft}\) has side length \(40 \div 4 = 10\,\text{ft}\), so its area is \(10^2 = 100\,\text{ft}^2\). This is greater than \(96\,\text{ft}^2\). 3. Let the rectangle's side lengths be \(x\) and \(y\). From \(2x + 2y = 40\), \(y = 20 - x\). 4. The area is \(A(x) = x(20 - x) = -x^2 + 20x\). Its vertex occurs at \(x = -\frac{20}{2 \cdot (-1)} = 10\). 5. Then \(y = 10\), so the maximum-area rectangle is a \(10\,\text{ft}\) by \(10\,\text{ft}\) square with area \(100\,\text{ft}^2\).

a) \(96\,\text{ft}^2\) b) Yes. The square has area \(100\,\text{ft}^2\). c) \(10\,\text{ft}\) by \(10\,\text{ft}\), with maximum area \(100\,\text{ft}^2\)

- At an intersection, set the two function values equal. - Factor the resulting quadratic equation. - Use the vertex formula to find each maximum height.

1. At an intersection, the heights are equal: \(-0.5x^2 + 2x + 1 = -x^2 + 3x + 1\). 2. Simplify: \(0.5x^2 - x = 0\). Factoring gives \(x(0.5x - 1) = 0\), so \(x = 0\) or \(x = 2\). The nonstarting intersection has \(x = 2\). 3. Evaluate either function: \(f(2) = -0.5 \cdot 2^2 + 2 \cdot 2 + 1 = 3\). The second intersection is \((2, 3)\). 4. For \(f\), the vertex occurs at \(x = -\frac{2}{2 \cdot (-0.5)} = 2\), and \(f(2) = 3\). Its vertex is \((2, 3)\), so its maximum height is \(3\,\text{m}\). 5. For \(g\), the vertex occurs at \(x = -\frac{3}{2 \cdot (-1)} = 1.5\), and \(g(1.5) = 3.25\). Its vertex is \((1.5, 3.25)\), so its maximum height is \(3.25\,\text{m}\). Therefore, the stream modeled by \(g\) reaches the greater maximum height.

a) \((2, 3)\) b) The stream modeled by \(g\). Its vertex is \((1.5, 3.25)\), compared with \((2, 3)\) for \(f\), so its maximum height is \(3.25\,\text{m}\) rather than \(3\,\text{m}\).

- Set the height equal to \(0\) to find the ground-level intercepts. - A centered vehicle extends half its width to each side of the y-axis. - For part c, find the two horizontal positions where the tunnel height equals the vehicle height.

1. At ground level, set \(h(x) = 0\): \(0 = -0.15x^2 + 5.4\). Then \(x^2 = 36\), so \(x = \pm 6\). The total ground-level width is \(12\,\text{m}\). 2. A centered \(3.20\,\text{m}\)-wide truck extends \(1.60\,\text{m}\) on either side of the center. At an outer edge, \(h(1.60) = -0.15 \cdot 1.60^2 + 5.4 = 5.016\,\text{m}\). 3. Since \(5.016\,\text{m} > 3.80\,\text{m}\), the truck can pass through the center. 4. For a vehicle height of \(4.50\,\text{m}\), solve \(4.50 = -0.15x^2 + 5.4\). Then \(0.15x^2 = 0.90\), so \(x^2 = 6\) and \(x = \pm\sqrt{6}\). 5. The greatest possible width is \(2\sqrt{6}\,\text{m} \approx 4.90\,\text{m}\).

a) \(12\,\text{m}\) b) Yes. The tunnel is \(5.016\,\text{m}\) high at the truck's outer edges, which exceeds \(3.80\,\text{m}\). c) \(2\sqrt{6}\,\text{m} \approx 4.90\,\text{m}\)

- Evaluate the function at \(x = 0\) to interpret the constant term. - Use the vertex formula to find the maximum. - Setting the function equal to \(10\) gives two distances; which one is on the descending side of the vertex?

1. Since \(h(0) = 6\), the ball is released from a height of \(6\,\text{ft}\). 2. The vertex occurs at \(x = -\frac{\frac{16}{15}}{2 \cdot \left(-\frac{4}{75}\right)} = 10\). Evaluate the height: \(h(10) = -\frac{4}{75} \cdot 10^2 + \frac{16}{15} \cdot 10 + 6 = \frac{34}{3}\). The maximum height is \(\frac{34}{3}\,\text{ft} \approx 11.33\,\text{ft}\). 3. Set the height equal to the hoop height: \(-\frac{4}{75}x^2 + \frac{16}{15}x + 6 = 10\). 4. Multiplying by \(75\) and simplifying gives \(x^2 - 20x + 75 = 0\), which factors as \((x - 5)(x - 15) = 0\). The ball is at hoop height at \(x = 5\) on the way up and at \(x = 15\) on the way down. 5. Therefore, the hoop must be \(15\,\text{ft}\) horizontally from the release point.

a) The ball is released from a height of \(6\,\text{ft}\). b) \(\frac{34}{3}\,\text{ft} \approx 11.33\,\text{ft}\) c) \(15\,\text{ft}\)

- Isolate \(t^2\) before taking the physically meaningful positive square root. - Keep the two calculations unrounded until the final requested hundredths. - For the factor comparison, compare the heights inside the square root rather than relying only on rounded times.

1. Solve the model for positive time: \(t=\sqrt{\frac{2h}{g}}\). 2. For \(h_1=45\,\text{m}\), \(t_1=\sqrt{\frac{2\cdot45}{9.81}}\,\text{s}\approx3.03\,\text{s}\). 3. For \(h_2=180\,\text{m}\), \(t_2=\sqrt{\frac{2\cdot180}{9.81}}\,\text{s}\approx6.06\,\text{s}\). 4. Exactly, \(\frac{t_2}{t_1}=\sqrt{\frac{180}{45}}=\sqrt{4}=2\). Multiplying the height by \(4\) doubles the fall time.

a) \(t_1\approx3.03\,\text{s}\); \(t_2\approx6.06\,\text{s}\) b) The fall time is multiplied by \(2\).

- Express the new price and the number of tickets sold using the same variable. - Revenue equals ticket price times tickets sold. - Factor the resulting quadratic equation. - Verify both possible price increases in the original revenue model.

1. Let \(x\) be the price increase in dollars. The new ticket price is \(40+x\), and the number of tickets sold is \(500-10x\). 2. Set revenue equal to the target: \((40+x)(500-10x)=20{,}210\). 3. Expand: \(20{,}000+100x-10x^2=20{,}210\). 4. Rearrange and divide by \(-10\): \(x^2-10x+21=0\). 5. Factor: \((x-3)(x-7)=0\), so \(x=3\) or \(x=7\). 6. Both values are meaningful: \(43\cdot470=20{,}210\) and \(47\cdot430=20{,}210\).

The ticket price could be increased by \(\$3\) or by \(\$7\).

- Write a growth factor for each year. - Remember that the second year begins with the first year's ending balance. - A one-percentage-point increase changes \(p\%\) to \((p+1)\%\). - Check whether each algebraic solution is meaningful as an interest rate.

1. The two annual growth factors are \(1+\frac{p}{100}\) and \(1+\frac{p+1}{100}\). 2. Model the final balance: \(2000\left(1+\frac{p}{100}\right)\left(1+\frac{p+1}{100}\right)=2226\). 3. Rewrite the factors and simplify: \(\frac{(p+100)(p+101)}{5}=2226\), so \((p+100)(p+101)=11{,}130\). 4. Expand and rearrange: \(p^2+201p-1030=0\). 5. Factor: \((p-5)(p+206)=0\), so \(p=5\) or \(p=-206\). 6. The value \(p=-206\) would represent a loss greater than \(100\%\) in the first year, which is not meaningful for this investment model. Therefore, \(p=5\).

\(5\%\)

- Translate the comparison between diagonals and vertices into an equation using the supplied formula. - Rearrange the resulting relation into quadratic standard form. - After solving, apply the geometric restriction on a polygon's number of vertices.

1. The condition “\(12\) more diagonals than vertices” gives \(\frac{n(n-3)}{2}=n+12\). 2. Multiply by \(2\) and rearrange: \(n^2-5n-24=0\). 3. Factor: \((n-8)(n+3)=0\), so \(n=8\) or \(n=-3\). 4. A polygon must have at least \(3\) vertices, so only \(n=8\) is valid.

The quadratic equation is \(n^2-5n-24=0\). Its roots are \(8\) and \(-3\), and only \(8\) is a valid polygon vertex count.

- Express the two vertex counts using one variable. - Apply the diagonal formula separately to each polygon before combining their counts. - Solve the resulting quadratic and check the geometric restriction on the smaller polygon.

1. The larger polygon has \(n+1\) vertices. 2. The total-diagonal condition is \(\frac{n(n-3)}{2}+\frac{(n+1)(n-2)}{2}=14\). 3. Simplify: \(2n^2-4n-2=28\), so \(n^2-2n-15=0\). 4. Factor: \((n-5)(n+3)=0\), giving \(n=5\) or \(n=-3\). 5. The smaller polygon must have at least \(3\) vertices, so \(n=5\). The polygons therefore have \(5\) and \(6\) vertices.

The quadratic equation is \(n^2-2n-15=0\). The valid solution is \(n=5\), so the polygons have \(5\) and \(6\) vertices.

- Write the diagonal formula once with \(n\) and once with \(2n\). - Represent the stated increase by subtracting the original count from the new count. - Simplify to a quadratic before solving and checking the valid domain for \(n\).

1. The original diagonal count is \(D(n)=\frac{n(n-3)}{2}\). 2. With \(2n\) vertices, the count is \(D(2n)=\frac{2n(2n-3)}{2}=n(2n-3)\). 3. The increase condition gives \(n(2n-3)-\frac{n(n-3)}{2}=18\). 4. Multiply by \(2\) and simplify: \(4n^2-6n-n^2+3n=36\), so \(n^2-n-12=0\). 5. Factor: \((n-4)(n+3)=0\), giving \(n=4\) or \(n=-3\). 6. Only \(n=4\) is a valid polygon vertex count.

The quadratic equation is \(n^2-n-12=0\), and the original polygon had \(4\) vertices.

- First determine the lawn area represented by \(62.5\%\) of the original park. - An inside path reduces each lawn dimension on two sides. - Decide the largest possible path width before solving the equation. - Use that domain to evaluate the algebraic roots.

1. The original area is \(120\cdot80=9600\,\text{ft}^2\), so the target lawn area is \(0.625\cdot9600=6000\,\text{ft}^2\). 2. The lawn dimensions are \(120-2x\) and \(80-2x\), so \((120-2x)(80-2x)=6000\). 3. Expanding and simplifying gives \(x^2-100x+900=0\). 4. The path must satisfy \(0<x<40\), because the smaller lawn dimension must remain positive. 5. Factoring gives \((x-10)(x-90)=0\), so \(x=10\) or \(x=90\). 6. Only \(x=10\) lies in the physical domain. The path width is \(10\,\text{ft}\).

a) \((120-2x)(80-2x)=6000\), equivalently \(x^2-100x+900=0\), with \(0<x<40\) b) \(10\,\text{ft}\)

- Subtract the border area from the full poster area to find the printed region's area. - How does a uniform border change both dimensions of the inner rectangle? - Reject a calculated width that would leave no printed region. - Compare the actual new border area with half of the original area.

1. The poster area is \(24\cdot16=384\,\text{in}^2\), so the printed region has area \(384-144=240\,\text{in}^2\). 2. The printed region has dimensions \(24-2x\) and \(16-2x\). Thus \((24-2x)(16-2x)=240\). 3. Expanding gives \(4x^2-80x+144=0\), or \((x-2)(x-18)=0\). 4. Only \(0<x<8\) is possible for a border, so \(x=2\,\text{in}\). 5. Half the width is \(1\,\text{in}\). The new interior area is \(22\cdot14=308\,\text{in}^2\), leaving border area \(384-308=76\,\text{in}^2\). 6. Since half the original border area would be \(72\,\text{in}^2\), the claim is false.

a) \(x=2\,\text{in}\) b) No. A \(1\,\text{in}\) border has area \(76\,\text{in}^2\), not \(72\,\text{in}^2\).

- Determine the base dimensions after the corner squares are removed. - The cutout side length becomes the box height. - Use the volume formula for a rectangular prism. - Check which algebraic solution can represent the original sheet.

1. Let \(a\) be the original side length in inches. 2. Cutting \(2\)-inch squares from both ends of each side leaves a square base with side length \(a-4\). The box height is \(2\,\text{in}\). 3. Use the volume: \(2(a-4)^2=72\). 4. Divide by \(2\): \((a-4)^2=36\). 5. Therefore, \(a-4=6\) or \(a-4=-6\), giving \(a=10\) or \(a=-2\). 6. A side length must be positive and greater than \(4\), so \(a=10\,\text{in}\).

\(10\,\text{in}\)

- Express the bed length in terms of its width. - A surrounding walkway adds its width on both sides of each dimension. - Use the outside area to form one equation in \(x\). - Check which root can represent a physical width.

1. Let \(x\) be the flower bed's width in feet. Its length is \(x+5\). 2. The walkway adds \(4\,\text{ft}\) to each overall dimension, so the outside dimensions are \(x+4\) and \(x+9\). 3. The area condition is \((x+4)(x+9)=150\), so \(x^2+13x-114=0\). 4. Factor: \((x-6)(x+19)=0\), so \(x=6\) or \(x=-19\). 5. A width must be positive, so the bed is \(6\,\text{ft}\) wide and \(11\,\text{ft}\) long.

a) \((x+4)(x+9)=150\), equivalently \(x^2+13x-114=0\) b) \(6\,\text{ft}\) wide and \(11\,\text{ft}\) long

- Use the perimeter to find the sum of the two side lengths. - Express one side length in terms of the other. - Use the area to form a quadratic equation. - Factor the resulting quadratic equation.

1. Let one side length be \(x\) inches. The perimeter gives a total side-length sum of \(13\), so the other side is \(13-x\). 2. Use the area: \(x(13-x)=40\). 3. Rearrange: \(x^2-13x+40=0\). 4. Factor: \((x-5)(x-8)=0\), so \(x=5\) or \(x=8\). 5. The corresponding other side is \(8\) or \(5\), so the rectangle's side lengths are \(5\,\text{in}\) and \(8\,\text{in}\).

\(5\,\text{in}\) and \(8\,\text{in}\)

- Write one product equation for each array arrangement. - Expanding the revised equation lets you use the original product without immediately dividing by a variable. - After eliminating one variable, check that the remaining equation is quadratic. - Reject any root that cannot represent a number of rows.

1. The original arrangement gives \(rb=400\). The revised arrangement gives \((r-4)(b+5)=360\). 2. Expand the revised equation: \(rb+5r-4b-20=360\). 3. Substitute \(rb=400\): \(5r-4b=-20\), so \(b=\frac{5r+20}{4}\). 4. Substitute into \(rb=400\): \(r\left(\frac{5r+20}{4}\right)=400\), which simplifies to \(r^2+4r-320=0\). 5. Factor: \((r-16)(r+20)=0\), so \(r=16\) or \(r=-20\). 6. A row count must be positive, so the original plan used \(16\) rows.

a) \(rb=400\) and \((r-4)(b+5)=360\) b) \(r^2+4r-320=0\) c) \(16\) rows

- Express the garden length using the width variable. - A surrounding path adds its width on both sides of each dimension. - Use the outside area to create one quadratic equation. - Interpret both roots in terms of the physical meaning of \(x\).

1. Let \(x\) be the garden's width in feet. Its length is \(2x\). 2. The path adds \(4\,\text{ft}\) to each overall dimension, so the outside dimensions are \(x+4\) and \(2x+4\). 3. The area condition is \((x+4)(2x+4)=160\), which simplifies to \(x^2+6x-72=0\). 4. Factor: \((x-6)(x+12)=0\), so \(x=6\) or \(x=-12\). 5. Width must be positive, so the garden is \(6\,\text{ft}\) wide and \(12\,\text{ft}\) long.

a) \((x+4)(2x+4)=160\), equivalently \(x^2+6x-72=0\) b) \(6\,\text{ft}\) wide and \(12\,\text{ft}\) long c) The root \(x=-12\) is rejected because a garden width cannot be negative.

- Use the vertex of the quadratic to find the minimum. - Convert each function value from hundreds of dollars to dollars. - Subtract the earlier cost from the later cost. - Compare the two first differences.

1. The parabola opens upward. Its vertex occurs at \(x=-\frac{b}{2a}=\frac{12}{1}=12\). 2. Evaluate \(K(12)=0.5\cdot 12^2-12\cdot 12+120=48\). Because the units are hundreds of dollars, the minimum cost is \(\$4800\). 3. Evaluate \(K(15)=52.5\) and \(K(14)=50\). The increase is \(2.5\) hundreds of dollars, or \(\$250\). 4. Evaluate \(K(19)=72.5\) and \(K(18)=66\). The increase is \(6.5\) hundreds of dollars, or \(\$650\). 5. The one-unit cost increase is larger farther to the right of the minimum. The cost graph becomes steeper as production increases in that region.

a) Minimum at \(x=12\); minimum cost \(\$4800\) b) \(\$250\) c) The increase from \(18\) to \(19\) is \(\$650\). Costs rise at an increasing rate to the right of the minimum.

- Substitute each speed directly into the model. - Find an increase by subtracting the earlier output from the later output. - Set the model equal to the given distance and solve the quadratic equation. - Reject solutions that are not meaningful speeds.

1. Evaluate \(s(30)=0.02\cdot 30^2+0.5\cdot 30=18+15=33\) feet. 2. Evaluate \(s(60)=0.02\cdot 60^2+0.5\cdot 60=72+30=102\) feet. 3. Compute \(s(50)=75\) and \(s(55)=88\), so the increase is \(13\) feet. 4. Set \(s(v)=75\): \(0.02v^2+0.5v-75=0\). Multiplying by \(50\) gives \(v^2+25v-3750=0\). 5. The quadratic formula gives \(v=\frac{-25\pm125}{2}\), so \(v=50\) or \(v=-75\). A negative speed is not meaningful in this context, so the answer is \(50\) mph.

a) \(33\) ft at \(30\) mph; \(102\) ft at \(60\) mph b) \(13\) ft c) \(50\) mph

- Find the positive zero of the quadratic in terms of \(k\). - Translate “does not travel past the edge” into an inequality. - The vertex lies halfway between the two zeros of a parabola.

1. The water returns to ground level at the positive zero of \(f_k\). Solve \(0.8x-\frac{k}{100}x^2=0\): \(x\left(0.8-\frac{k}{100}x\right)=0\). The positive zero is \(x=\frac{80}{k}\). 2. The water must land no farther than \(16\,\text{ft}\) from the nozzle, so \(\frac{80}{k}\le16\). Because \(k>0\), this gives \(k\ge5\). The smallest possible value is \(k=5\). 3. For \(k=5\), the zeros are \(0\) and \(16\), so the vertex lies halfway between them at \(x=8\). 4. Evaluate the function at the vertex: \(f_5(8)=0.8\cdot 8-0.05\cdot 8^2=6.4-3.2=3.2\).

a) \(k=5\) b) \(3.2\,\text{ft}\)

- Use the given horizontal range to write an equation involving \(c\). - Substitute the obstacle's horizontal position into the function. - Compare the resulting height with the obstacle's height.

1. Returning to the ground at \(x=50\) means \(h_c(50)=0\). Thus, \(50c-0.01\cdot 50^2=0\). 2. Simplify: \(50c-25=0\), so \(c=0.5\). 3. Evaluate the height at the obstacle: \(h_{0.5}(25)=0.5\cdot 25-0.01\cdot 25^2=12.5-6.25=6.25\). 4. Since \(6.25<7\), the projectile does not clear the obstacle.

a) \(c=0.5\) b) No. Its height at \(x=25\) is \(6.25\,\text{ft}\), which is less than \(7\,\text{ft}\).

- Choose an exact point on each graph. - Substitute the point's speed and force into \(F = av^2\). - For “percent greater,” use the racing-position force as the reference value.

1. For the upright position, the graph includes the point \((10, 40)\). Substitute into \(F = av^2\): \(40 = a \cdot 10^2\), so \(a = 0.4\). 2. For the racing position, the graph includes the point \((10, 25)\). Then \(25 = a \cdot 10^2\), so \(a = 0.25\). 3. At \(10\,\text{m/s}\), the difference is \(40 - 25 = 15\,\text{N}\). Relative to the racing-position value, the percent increase is \(\frac{15}{25} \cdot 100\% = 60\%\).

a) Upright position: \(a = 0.4\) Racing position: \(a = 0.25\) b) \(60\%\) greater

- Compare the graph's y-intercept, opening direction, and vertex with each option. - Set the correct function equal to the hoop height and choose the solution after the vertex. - For each shot, compare the vertex height and the positive zero.

1. The graph starts at \((0, 6)\), opens downward, and has vertex \((10, 11)\). Function (1) has \(f(0) = 6\), opens downward, and has vertex at \(x = -\frac{1}{2 \cdot (-0.05)} = 10\), with \(f(10) = 11\). Therefore, function (1) matches the graph. 2. Set the first function equal to the hoop height: \(-0.05x^2 + x + 6 = 10\). This is equivalent to \(x^2 - 20x + 80 = 0\), so \(x = 10 \pm 2\sqrt{5}\). The descending solution is \(x = 10 + 2\sqrt{5} \approx 14.47\,\text{ft}\). 3. The first ball has maximum height \(11\,\text{ft}\). Its positive ground-level intercept solves \(-0.05x^2 + x + 6 = 0\), giving \(x = 10 + 2\sqrt{55} \approx 24.83\,\text{ft}\). 4. For the second ball, the vertex occurs at \(x = -\frac{0.8}{2 \cdot (-0.04)} = 10\), and \(k(10) = 10\,\text{ft}\). Its positive ground-level intercept solves \(-0.04x^2 + 0.8x + 6 = 0\), giving \(x = 10 + 5\sqrt{10} \approx 25.81\,\text{ft}\). 5. The second ball does not travel higher, but it travels farther.

a) Function (1), \(f(x) = -0.05x^2 + x + 6\) b) \(10 + 2\sqrt{5}\,\text{ft} \approx 14.47\,\text{ft}\) c) The second ball travels lower but farther. Its maximum height is \(10\,\text{ft}\), compared with \(11\,\text{ft}\) for the first; its range is about \(25.81\,\text{ft}\), compared with about \(24.83\,\text{ft}\).

- Use two points on the revenue line to find its slope. - Break-even occurs where revenue equals cost. - Factor the resulting quadratic equation. - Profit equals revenue minus cost.

1. The revenue graph is a line through \((0, 0)\) and \((10, 400)\). Its slope is \(\frac{400-0}{10-0}=40\), so \(E(x)=40x\). 2. At a break-even point, revenue equals cost: \(40x=0.5x^2+15x+200\). 3. Rearrange and multiply by \(2\): \(x^2-50x+400=0\). 4. Factor: \((x-10)(x-40)=0\), so the break-even production levels are \(x=10\) and \(x=40\). 5. At \(x=30\), the revenue is \(E(30)=40\cdot30=1200\), or \(\$1200\). 6. The cost is \(K(30)=0.5\cdot30^2+15\cdot30+200=1100\), or \(\$1100\). 7. The profit is \(\$1200-\$1100=\$100\).

a) \(E(x)=40x\) b) \(10\) bottles and \(40\) bottles c) \(\$100\)

- Substitute a clearly readable point from each curve into the quadratic model. - Keep the scale factor \(\frac{v}{10}\) together before squaring. - For the percent comparison, identify which curve supplies the reference value.

1. Curve \(f\) contains the point \((20,8)\). Substituting gives \(8=k\left(\frac{20}{10}\right)^2=4k\), so \(k_f=2\). 2. Curve \(g\) contains the point \((20,4)\). Substituting gives \(4=4k\), so \(k_g=1\). 3. At \(v=25\), \(F_f=2(2.5)^2=12.5\,\text{kN}\), and \(F_g=(2.5)^2=6.25\,\text{kN}\). 4. The percent increase from \(g\) to \(f\) is \(\frac{12.5-6.25}{6.25}\cdot100\%=100\%\).

a) Curve \(f\): \(k=2\); curve \(g\): \(k=1\) b) The force on curve \(f\) is \(100\%\) greater.

- Use two exact points on the linear graph to find its slope. - At an intersection, the two plans have equal balances. - Convert \(3\) years to months before evaluating \(s_1\).

1. The linear graph for \(s_2\) passes through \((0, 0)\) and \((30, 450)\). Its slope is \(\frac{450}{30} = 15\), so \(s_2(x) = 15x\). 2. Set the balances equal: \(0.5x^2 = 15x\). Factoring gives \(x(0.5x - 15) = 0\), so \(x = 0\) or \(x = 30\). After the start, the plans next have the same balance at \(30\) months. 3. Three years is \(36\) months. Plan \(s_1\) gives \(s_1(36) = 0.5 \cdot 36^2 = 648\). Since \(\$648 > \$600\), the purchase is affordable, with \(\$48\) left over.

a) \(s_2(x) = 15x\) b) \(30\) months c) Yes. The balance is \(\$648\), so \(\$48\) remains.

- Which graph has an increasing rate of change? - Use an exact point on the line to find its slope. - At the given speed, read both component distances before adding them.

1. Braking distance grows quadratically, so it is represented by the curved graph \(p\). 2. The linear graph \(k\) passes through \((0, 0)\) and \((50, 60)\). Its slope is \(\frac{60}{50}=1.2\), so \(k(v)=1.2v\). 3. At \(50\,\text{mph}\), the reaction distance is \(k(50)=1.2\cdot50=60\,\text{ft}\). 4. From graph \(p\), the braking distance at \(50\,\text{mph}\) is \(100\,\text{ft}\). 5. The total stopping distance is \(60+100=160\,\text{ft}\).

a) Graph \(p\), because it has quadratic growth b) \(k(v)=1.2v\) c) \(160\,\text{ft}\)

- Use the origin and another clear point on line \(a\) to find its slope. - Equal costs occur where the two graphs intersect. - Verify the positive intersection by setting the functions equal. - Substitute \(50\) into both functions and compare.

1. The graph for machine A is a line through the origin and \((40, 480)\). Its slope is \(\frac{480}{40}=12\), so \(A(x)=12x\). 2. Set the costs equal: \(0.3x^2=12x\). 3. Since \(x>0\), divide by \(x\): \(0.3x=12\), so \(x=40\). 4. At \(x=40\), both machines cost \(\$480\). 5. At \(x=50\), machine A costs \(A(50)=12\cdot50=600\), or \(\$600\). 6. Machine B costs \(B(50)=0.3\cdot50^2=750\), or \(\$750\). 7. Machine A is less expensive by \(\$150\) per day.

a) \(A(x)=12x\) b) The costs are equal at \(x=40\) units, when each costs \(\$480\). c) The company would save \(\$150\) per day by choosing machine A.

- Revenue is zero when nothing is sold; look for the graph that passes through the origin. - Fixed cost is read where the cost graph meets the y-axis. - Profit is positive where its graph lies above the x-axis. - The maximum profit is represented by the vertex of the profit parabola.

1. Graph \(f\) is revenue because it is a line through the origin. Graph \(g\) is total cost because it has a positive y-intercept representing fixed cost. Graph \(h\) is profit because it is negative when costs exceed revenue and positive between its two x-intercepts. 2. Graph \(g\) has y-intercept \(10\). Since the vertical scale is hundreds of dollars, the fixed cost is \(\$1000\). 3. Graph \(h\) crosses the x-axis at about \(x=3\) and \(x=17\). Profit is positive between those intercepts, corresponding to about \(300\) to \(1700\) gallons. 4. The vertex of graph \(h\) is \((10,10)\). Therefore, the maximum profit is \(\$1000\) at a production quantity of \(1000\) gallons.

a) \(f\): revenue, \(g\): total cost, \(h\): profit b) \(\$1000\) c) About \(300\) to \(1700\) gallons d) A maximum profit of \(\$1000\) at \(1000\) gallons

- Use the vertex formula to find the horizontal coordinate of the maximum. - Ground contact means the height is \(0\). - Keep only the root that represents a positive horizontal distance.

1. The vertex occurs at \(x = -\frac{0.6}{2 \cdot (-0.05)} = 6\). 2. Evaluate the height: \(h(6) = -0.05 \cdot 6^2 + 0.6 \cdot 6 + 2 = 3.8\). The vertex is \((6, 3.8)\). This means the shot reaches its maximum height of \(3.8\,\text{m}\) after traveling \(6\,\text{m}\) horizontally. 3. Set the height equal to \(0\): \(-0.05x^2 + 0.6x + 2 = 0\). Dividing by \(-0.05\) gives \(x^2 - 12x - 40 = 0\). 4. Using the quadratic formula, \(x = 6 \pm \sqrt{76}\). The context-valid solution is \(x = 6 + \sqrt{76} \approx 14.72\,\text{m}\).

a) \((6, 3.8)\). The shot reaches its maximum height of \(3.8\,\text{m}\) after \(6\,\text{m}\) of horizontal travel. b) \(6 + \sqrt{76}\,\text{m} \approx 14.72\,\text{m}\)

- First evaluate the model exactly at the proposed input. - Compare the resulting output with what is possible for a visitor count. - Compare the proposed temperature with the interval on which the model was fitted.

1. Substitute \(T=20\): \(A(20)=-2(20-70)^2+900\). 2. This gives \(A(20)=-2(2500)+900=-4100\). 3. A prediction of \(-4100\) visitors is not physically meaningful. 4. Also, \(20^\circ\text{F}\) lies far outside the observed interval from \(50^\circ\text{F}\) to \(90^\circ\text{F}\). The model should not be trusted for that extrapolation.

The model gives \(A(20)=-4100\), which is not a meaningful attendance count. The prediction should not be trusted because \(20^\circ\text{F}\) is far outside the temperature range used to fit the model.

- Begin with a factored quadratic containing an unknown scale factor and the two break-even inputs. - Use the given profit at \(x=6\) to determine that factor. - Profit is positive only on the part of the between-roots interval that also lies inside the feasible domain. - Evaluate the model at the capacity endpoint before comparing it with the given profit at \(x=6\).

1. The break-even points give \(P(x)=a(x-2)(x-10)\). 2. Use \(P(6)=6400\): \(6400=a(4)(-4)=-16a\), so \(a=-400\). 3. Thus, \(P(x)=-400(x-2)(x-10)\). 4. The model is positive between its zeros, \(2<x<10\). Intersecting with the feasible domain \(0\le x\le8\) gives \(2<x\le8\), corresponding to more than \(200\) and at most \(800\) units. 5. At full capacity, \(P(8)=-400(6)(-2)=4800\), so the modeled profit is \(\$4800\). 6. Since \(P(6)=\$6400\), increasing sales from \(600\) to \(800\) units lowers the modeled profit by \(\$1600\).

a) \(P(x)=-400(x-2)(x-10)\) b) \(2<x\le8\), meaning more than \(200\) and at most \(800\) units c) \(P(8)=\$4800\); increasing sales from \(600\) to \(800\) units lowers the modeled profit by \(\$1600\).

- Compare the vertices of the two curves first. - Use the grid to read both outputs at the specified fertilizer amount. - An equal prediction occurs where the two curves intersect. - After the positive intersection, compare which curve lies above the other.

1. From the graph, \(f\) has vertex \((6,18)\), while \(g\) has vertex \((8,16)\). Thus, \(f\) has the higher maximum: \(18\) pounds at \(6\) ounces; \(g\) reaches \(16\) pounds at \(8\) ounces. 2. At \(4\) ounces, the graph gives \(f(4)=16\) and \(g(4)=12\), so \(f\) is higher by \(4\) pounds. 3. Besides the origin, the curves intersect at \((8,16)\). Thus, the positive fertilizer amount at which the predictions are equal is \(8\) ounces. 4. For fertilizer amounts greater than \(8\) ounces through the displayed endpoint at \(12\) ounces, graph \(g\) lies above graph \(f\).

a) \(f\) has the higher maximum: \(18\) pounds at \(6\) ounces. Model \(g\) reaches \(16\) pounds at \(8\) ounces. b) At \(4\) ounces, \(f\) is higher by \(4\) pounds. c) The positive intersection is at \(8\) ounces. After that point, \(g\) is higher through the displayed range.

- Do not add signed residuals directly; positive and negative errors can cancel. - Square each residual before comparing the total error for the two models. - Beyond error size, inspect whether the signs follow a visible pattern across increasing \(x\). - A useful model should leave relatively small residuals without a strong systematic shape.

1. For the linear model, the sum of squared residuals is \(2^2+(-1)^2+(-2)^2+(-1)^2+2^2=14\). 2. For the quadratic model, the sum of squared residuals is \(0.5^2+(-0.5)^2+0^2+0.5^2+(-0.5)^2=1\). 3. The linear residuals are positive at both ends and negative through the middle, showing a curved systematic pattern rather than random scatter around zero. 4. The quadratic residuals are much smaller and alternate around zero without the same systematic curvature. 5. The quadratic model is the better fit because it has the much smaller sum of squared residuals and leaves no comparable systematic pattern.

a) Linear model: \(14\); quadratic model: \(1\) b) The linear residuals show a systematic positive-at-the-ends, negative-in-the-middle pattern. The quadratic residuals are small and lie on both sides of zero without that curved pattern. c) The quadratic model is better because its squared-residual total is much smaller and its residuals show less systematic structure.

- A residual is represented visually by the vertical difference between a measured point and the model at the same horizontal input. - Compare vertical gaps, not horizontal distances from the curve. - Use whether the point lies above or below the curve to determine the residual's sign. - After setting aside the largest discrepancy, examine both the size and the direction of the remaining deviations.

1. Compare each measured point with the model curve at the same horizontal distance. Point \(D\) has by far the largest vertical separation from \(q\). 2. At \(D\), the measured height is about \(10\,\text{ft}\), while the model predicts about \(16\,\text{ft}\). The residual is therefore about \(-6\,\text{ft}\). 3. The residual is negative because \(D\) lies below the model curve. 4. After removing \(D\), each remaining point lies within about \(1\,\text{ft}\) of \(q\), with some points above and some below the curve. 5. Thus, the remaining measurements support \(q\) as a reasonable model over the displayed interval; they show small residuals without a one-sided pattern.

a) \(D\) b) Negative, approximately \(-6\,\text{ft}\) c) Yes. After removing \(D\), the other six points are all within about \(1\,\text{ft}\) of the model and occur on both sides of the curve, so their residuals are small without a systematic one-sided pattern.

- Choose a variable for the number of price changes, not for the revenue itself. - How do the price and ticket count each change when the variable increases by one? - Multiply price by the expected number of sales before expanding.

1. Let \(x\) be the number of \(\$1\) price increases. 2. Then the price is \(10+x\) dollars and the expected number of tickets is \(150-10x\). 3. Revenue is price times tickets: \(R(x)=(10+x)(150-10x)\). 4. Expanding gives \(R(x)=-10x^2+50x+1500\).

For example, let \(x\) be the number of \(\$1\) price increases. Then \(R(x)=-10x^2+50x+1500\) dollars.

- Which table value gives the constant term immediately? - Use the other two rows to find equations for the two remaining coefficients. - Check the completed rule against all the costs.

1. Since \(C(0)=6\), the constant term is \(c=6\). 2. Using \(C(1)=12\) gives \(a+b=6\). Using \(C(2)=22\) gives \(4a+2b=16\), or \(2a+b=8\). 3. Subtracting the equations gives \(a=2\), and then \(b=4\). 4. Therefore, \(C(n)=2n^2+4n+6\). This matches all three table values.

\(C(n)=2n^2+4n+6\)

- Work out each model's predictions at the times in the table. - A match at one time is not enough. Check all five times before choosing. - What limits must a count of people satisfy, especially when predicting beyond the table?

1. The quadratic gives \(Q(0)=5\), \(Q(1)=11\), \(Q(2)=13\), \(Q(3)=11\), and \(Q(4)=5\), matching every table value. 2. The linear model gives \(L(1)=7\) and \(L(4)=13\), which do not match the table; choose \(Q\). 3. Extrapolating gives \(Q(6)=-2(6-2)^2+13=-19\). 4. A negative number of people is impossible, and \(t=6\) is beyond the simulated \(0\)-to-\(4\)-minute period. The prediction should not be trusted.

a) \(Q\) matches all five values: \(Q(0)=5\), \(Q(1)=11\), \(Q(2)=13\), \(Q(3)=11\), and \(Q(4)=5\). The other model fails, for example \(L(1)=7\) instead of \(11\). b) \(Q(6)=-19\); it is not a sensible prediction because it is negative and beyond the simulation's time range.

- Use the three data points to create a system for \(a\), \(b\), and \(c\). - How can subtracting equations reduce the system? - What height represents landing? - Which quadratic-equation solution is meaningful as time?

1. Let \(h(t) = at^2 + bt + c\). Substituting the three data points gives \(a + b + c = 15\), \(4a + 2b + c = 18\), and \(9a + 3b + c = 17\). 2. Subtract consecutive equations: \(3a + b = 3\) and \(5a + b = -1\). Their difference gives \(2a = -4\), so \(a = -2\). Then \(b = 9\), and \(c = 8\). 3. The model is \(h(t) = -2t^2 + 9t + 8\). 4. The initial height is \(h(0) = 8\,\text{m}\). 5. Solve \(-2t^2 + 9t + 8 = 0\): \(t = \frac{9 \pm \sqrt{145}}{4}\). The positive solution is \(t \approx 5.26\), so the drone reaches the ground after about \(5.26\,\text{s}\).

a) \(h(t) = -2t^2 + 9t + 8\) b) \(8\,\text{m}\) c) About \(5.26\,\text{s}\)

- Use the maximum and the two points where the arch meets the road. - The truck's full width matters, not just the clearance above its center. - Which horizontal positions give the lowest clearance over the truck? - Compare that clearance with the truck's height.

1. The vertex is \((0,10)\), so \(c=10\). The road-level zeros are at \(x=-20\) and \(x=20\). 2. Substituting \((20,0)\) gives \(0=400a+10\), so \(a=-0.025\) and \(f(x)=-0.025x^2+10\). 3. At \(x=15\), the arch height is \(f(15)=-0.025\cdot 15^2+10=4.375\,\text{ft}\). 4. The centered truck's edges lie at \(x=-3\) and \(x=3\). The lowest clearance across its width is \(f(3)=10-0.025\cdot9=9.775\,\text{ft}\). 5. Since \(9.775\,\text{ft}>9.5\,\text{ft}\), the truck fits in this model.

a) \(f(x)=-0.025x^2+10\) b) \(4.375\,\text{ft}\) c) Yes. The clearance at the truck's outer edges is \(9.775\,\text{ft}\), which is more than \(9.5\,\text{ft}\).

- Split the left-hand fraction into two terms. - How is \(\frac{b}{a}\) related to \(\delta\)? - Check which algebraic root matches the stated lengths.

1. Rewrite \(\frac{2a+b}{a}=2+\frac{b}{a}=2+\frac{1}{\delta}\). 2. Thus \(2+\frac{1}{\delta}=\delta\). Multiplying by \(\delta\) gives \(\delta^2-2\delta-1=0\). 3. The quadratic formula gives \(\delta=\frac{2\pm\sqrt{4+4}}{2}=1\pm\sqrt{2}\). 4. Since \(a\) and \(b\) are positive lengths, \(\delta>0\); reject \(1-\sqrt{2}<0\).

a) \(\delta^2-2\delta-1=0\) b) \(\delta=1+\sqrt{2}\). A ratio of positive lengths cannot be negative.

- Use the maximum point to write vertex form. - Substitute the starting point to find \(a\). - If the new range is doubled, where is the second zero? - The vertex lies midway between the zeros.

1. Use vertex form with vertex \((2, 4)\): \(f(x) = a(x - 2)^2 + 4\). 2. Since the stream passes through \((0, 0)\), \(0 = a(0 - 2)^2 + 4\), so \(4a = -4\) and \(a = -1\). 3. Expand: \(f(x) = -(x - 2)^2 + 4 = -x^2 + 4x\). 4. Factor \(f(x) = -x(x - 4)\). The zeros are \(x = 0\) and \(x = 4\), so the water returns to the surface \(4\,\text{m}\) from the outlet. 5. Doubling the range places the second zero at \(x = 8\). With the same leading coefficient, the new function is \(g(x) = -x(x - 8) = -x^2 + 8x\). 6. The new vertex lies midway between the zeros at \(x = 4\). Its height is \(g(4) = -4^2 + 8 \cdot 4 = 16\). The maximum height increases from \(4\,\text{m}\) to \(16\,\text{m}\), so it is quadrupled.

a) \(f(x) = -x^2 + 4x\) b) \(4\,\text{m}\) c) The maximum height quadruples, increasing from \(4\,\text{m}\) to \(16\,\text{m}\).

- Factor out \(x\) to find each pair of zeros. - The vertex lies midway between the zeros. - Write the nonzero zero and the vertex height in terms of \(a\) and \(b\).

1. For \(f\), the vertex occurs at \(x = -\frac{2}{2 \cdot (-0.05)} = 20\). Then \(f(20) = -0.05 \cdot 20^2 + 2 \cdot 20 = 20\), so the maximum height is \(20\,\text{m}\). 2. Factor \(f(x) = x(-0.05x + 2)\). The zeros are \(x = 0\) and \(x = 40\), so the range is \(40\,\text{m}\). 3. For \(g\), the vertex occurs at \(x = -\frac{4}{2 \cdot (-0.05)} = 40\). Then \(g(40) = -0.05 \cdot 40^2 + 4 \cdot 40 = 80\), so the maximum height is \(80\,\text{m}\). 4. Factor \(g(x) = x(-0.05x + 4)\). The zeros are \(x = 0\) and \(x = 80\), so the range is \(80\,\text{m}\). 5. In general, the nonzero zero of \(ax^2 + bx = x(ax + b)\) is \(x = -\frac{b}{a}\). With \(a\) fixed, doubling \(b\) doubles the range. 6. The vertex height is \(-\frac{b^2}{4a}\). With \(a < 0\) fixed, doubling \(b\) multiplies the maximum height by \(2^2 = 4\).

a) For \(f\): maximum height \(20\,\text{m}\), range \(40\,\text{m}\) For \(g\): maximum height \(80\,\text{m}\), range \(80\,\text{m}\) b) Doubling \(b\) doubles the range and quadruples the maximum height when \(a\) remains fixed and negative.

- Draw the two plots and count the two outer side fences plus the shared divider. - Write the fencing constraint using the depth and total width. - Express the total area as a quadratic in one variable. - Compare the two maximizing dimensions to decide whether the layout is square.

1. Let \(x\) be the depth of the layout. Three fence segments have length \(x\): the two outer sides and the divider. Let \(y\) be the total width, so two outer fence segments have length \(y\). 2. The fencing constraint is \(3x + 2y = 120\), so \(y = 60 - 1.5x\). 3. The total area is \(A(x) = xy = x(60 - 1.5x) = -1.5x^2 + 60x\). 4. The maximum occurs at the vertex: \(x = -\frac{60}{2 \cdot (-1.5)} = 20\). Then \(y = 60 - 1.5 \cdot 20 = 30\). 5. The maximum total area is \(20 \cdot 30 = 600\,\text{ft}^2\). Since \(20 \neq 30\), the entire layout is not a square.

The maximizing layout is \(20\,\text{ft}\) by \(30\,\text{ft}\), with total area \(600\,\text{ft}^2\). It is not a square.

- Use the two ground-level supports as the zeros of the quadratic. - Substitute the known clearance point to determine the leading coefficient. - Find the horizontal positions of the truck's outer edges and evaluate the arch height there.

1. Place the left support at \((0, 0)\) and the right support at \((12, 0)\). Write the arch in intercept form: \(f(x) = ax(x - 12)\). 2. Use the point \((2, 4)\): \(4 = a \cdot 2 \cdot (2 - 12) = -20a\), so \(a = -0.2\). Therefore, \(f(x) = -0.2x(x - 12) = -0.2x^2 + 2.4x\). 3. The bridge center is at \(x = 6\). A centered \(3.2\,\text{m}\)-wide truck extends from \(x = 6 - 1.6 = 4.4\) to \(x = 6 + 1.6 = 7.6\). 4. The lowest clearance above the truck occurs at its outer edges. By symmetry, evaluate one edge: \(f(4.4) = -0.2 \cdot 4.4^2 + 2.4 \cdot 4.4 = 6.688\,\text{m}\). 5. Since \(6.688\,\text{m} > 4.5\,\text{m}\), the truck can pass beneath the bridge while centered.

Yes. The clearance at the truck's outer top corners is \(6.688\,\text{m}\), which is greater than the truck's height of \(4.5\,\text{m}\).

- Represent each area as length times width, once before the change and once after it. - Use one equation to express one original dimension in terms of the other. - After substitution, clear any denominator and simplify until the required quadratic appears. - Check the roots against the fact that both dimensions must be positive.

1. Let \(l\) be the original length and \(w\) the original width, in feet. The two area conditions are \(lw=1500\) and \((l+10)(w-5)=1500\). 2. From \(lw=1500\), write \(w=\frac{1500}{l}\). Substitute into the changed-area equation: \((l+10)\left(\frac{1500}{l}-5\right)=1500\). 3. Multiply by \(l\) and simplify: \((l+10)(1500-5l)=1500l\), which gives \(l^2+10l-3000=0\). 4. Factor: \((l-50)(l+60)=0\). Thus, \(l=50\) or \(l=-60\). 5. A length must be positive, so the valid root is \(l=50\). Then \(w=\frac{1500}{50}=30\).

a) \(lw=1500\) and \((l+10)(w-5)=1500\) b) One valid eliminated equation is \(l^2+10l-3000=0\), with roots \(l=50\) and \(l=-60\). The physically valid root is \(l=50\). c) The original lot was \(50\,\text{ft}\) long and \(30\,\text{ft}\) wide.

- Model each arrangement as number of rows times spaces per row. - Use the original-total equation to express spaces per row in terms of the number of rows. - Substitute into the renovated-total equation and simplify to the requested quadratic. - After solving, translate each original row count to the corresponding renovated row count and check both arrangements.

1. Let \(r\) be the original number of rows and \(s\) the original number of spaces per row. The two arrangements give \(rs=300\) and \((r+5)(s+2)=440\). 2. From \(rs=300\), write \(s=\frac{300}{r}\). Substitute into the renovated-arrangement equation: \((r+5)\left(\frac{300}{r}+2\right)=440\). 3. Multiply by \(r\) and simplify: \((r+5)(300+2r)=440r\), so \(r^2-65r+750=0\). 4. Factor: \((r-15)(r-50)=0\). Thus, \(r=15\) or \(r=50\). 5. After adding \(5\) rows, the renovated lot could have \(20\) rows or \(55\) rows. 6. If \(r=15\), then \(s=20\), so the renovated arrangement is \(20\) rows of \(22\) spaces, giving \(20\cdot22=440\). If \(r=50\), then \(s=6\), so the renovated arrangement is \(55\) rows of \(8\) spaces, giving \(55\cdot8=440\). Both configurations satisfy all stated conditions.

a) \(rs=300\) and \((r+5)(s+2)=440\) b) \(r^2-65r+750=0\), so \(r=15\) or \(r=50\). c) The renovated parking lot could have \(20\) rows or \(55\) rows. Both answers satisfy the original and renovated space totals.

- Cutting a square from each corner changes both base dimensions by twice the cut size. - Express the folded box dimensions using the original width variable. - What restriction must the original width satisfy so the base dimensions remain positive? - Compare each algebraic root with that restriction.

1. Let \(x\) be the original width in centimeters. The original length is \(x+10\). 2. After the corner squares are removed, the box is \(2\,\text{cm}\) high and has base dimensions \(x-4\) and \(x+6\). The cardboard must satisfy \(x>4\). 3. The volume condition is \(2(x-4)(x+6)=192\). 4. Dividing by \(2\) and expanding gives \(x^2+2x-120=0\). 5. Factor: \((x-10)(x+12)=0\), so \(x=10\) or \(x=-12\). 6. Only \(x=10\) satisfies the physical domain. The original width is \(10\,\text{cm}\), and the original length is \(20\,\text{cm}\).

a) \(2(x-4)(x+6)=192\), equivalently \(x^2+2x-120=0\), with \(x>4\) b) \(20\,\text{cm}\) long and \(10\,\text{cm}\) wide

- a) Substitute \(90\) for \(d\) and solve the resulting quadratic equation. - b) Replace \(n\) with \(n+1\) in the diagonal formula. - Subtract the original formula from the new formula. - Consider why adding a vertex also changes one former side into a diagonal.

1. For part a), set the diagonal formula equal to \(90\): \(\frac{n(n-3)}{2}=90\). 2. Rearrange: \(n^2-3n-180=0\). 3. Factor: \((n-15)(n+12)=0\), so \(n=15\) or \(n=-12\). 4. A polygon cannot have a negative number of vertices, so \(n=15\). 5. For part b), a polygon with \(n+1\) vertices has \(d_{n+1}=\frac{(n+1)(n-2)}{2}\) diagonals. 6. Find the increase: \(d_{n+1}-d_n=\frac{(n+1)(n-2)-n(n-3)}{2}=n-1\). 7. For \(n=15\), the increase is \(15-1=14\) diagonals.

a) \(15\) vertices b) The increase is \(n-1\) diagonals. For the polygon in part a), the increase is \(14\) diagonals.

- a) Subtract the photo area from the outside rectangle's area. - b) Expand the general frame-area expression before substituting dimensions. - Compare the sums of the two photos' side lengths. - The two photos have equal area, but they do not have equal perimeters.

1. For part a), the frame area is \((4+2x)(6+2x)-4\cdot6\). Set it equal to \(24\): \((4+2x)(6+2x)-24=24\). 2. Expand and simplify: \(4x^2+20x-24=0\), so \(x^2+5x-6=0\). 3. Factor: \((x-1)(x+6)=0\). Since a width must be positive, \(x=1\,\text{in}\). 4. For a rectangle with side lengths \(a\) and \(b\), a frame of width \(x\) has area \((a+2x)(b+2x)-ab=2x(a+b)+4x^2\). 5. The first photo has \(a+b=4+6=10\), while the second has \(a+b=3+8=11\). 6. With the same positive width \(x\), the larger value of \(a+b\) gives a larger frame area. In fact, the second frame area is \((3+2)(8+2)-24=26\,\text{in}^2\).

a) \(1\,\text{in}\) b) The second frame area is greater. It is \(26\,\text{in}^2\), because the second photo has the larger sum of side lengths.

- Find the amount and fraction of vinegar after the first replacement. - The second removal contains the same vinegar fraction as the mixed tank. - Write an expression for the vinegar remaining after both removals. - Check the possible values against the tank capacity.

1. Let \(x\) be the number of gallons removed each time. 2. After the first removal and replacement, \(18-x\) gallons of vinegar remain, so the vinegar fraction is \(\frac{18-x}{18}\). 3. During the second removal, the amount of vinegar removed is \(x\left(\frac{18-x}{18}\right)\). 4. The vinegar remaining is \((18-x)-x\left(\frac{18-x}{18}\right)=\frac{(18-x)^2}{18}\). 5. Set this equal to \(8\): \(\frac{(18-x)^2}{18}=8\), so \((18-x)^2=144\). 6. Thus, \(18-x=12\) or \(18-x=-12\), giving \(x=6\) or \(x=30\). 7. The amount removed cannot exceed the tank capacity, so \(x=6\) gallons.

\(6\) gallons were removed each time.