Quadratic Optimization: Free Response

1. Two vertices, two different questions . Foundational, 9 points. Question 1 of 5.

Two quadratic quantities are given: $y=2x^2-8x+11$ and $y=-x^2+14x-40$ . One part below asks for a VALUE; another asks for an INPUT. Read carefully and answer exactly what is asked, no more and no less.

Part A.
The quantity $y=2x^2-8x+11$ has a minimum. Find the minimum value.

Solve and show your work Write each step out, and end with the value and its units. 3 points
Part B.
The quantity $y=-x^2+14x-40$ has a maximum. Find the optimizing input, that is, the value of $x$ at which the maximum occurs, not the maximum value itself.

Solve and show your work Write each step out, and end with the value and its units. 3 points
Part C.
Explain why finding the optimizing input in part B does not, by itself, give you the maximum value of that same quantity, and name the one additional step that would.

Explain why it works A sentence or two. Reasons, not steps. 3 points

Hints

One at a time, each one a step further than the last. Take only as many as you need.

Hint 1 of 3
Every vertex has two coordinates, an input and a value, and a question about one is not automatically answered by finding the other. Work out which coordinate each part is actually asking for before you compute anything.
Hint 2 of 3 · Part A
For part A, use $x=-\tfrac{b}{2a}$ to find where the minimum happens, then substitute that number back into the equation. What the question wants is what comes out of that substitution.
Hint 3 of 3 · Part B
For part B, the phrase optimizing input tells you exactly which vertex coordinate to report. Stop the moment you have it: this part does not ask you to substitute it back in.

Answer and solution

Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Uses $x=-\tfrac{b}{2a}$ to locate the vertex input, then substitutes that input back into the equation. . Worth 2 points.

Reports specifically the minimum VALUE, not the input at which it occurs. . Worth 1 point.

Part B 3 points

Correctly computes the vertex input using $a=-1$ and $b=14$ , handling both signs correctly. . Worth 2 points.

Reports specifically the optimizing INPUT, not the value the quantity reaches there. . Worth 1 point.

Part C 3 points

Explains that a vertex has two separate coordinates, so locating the input in part B fixes only one of them. . Worth 2 points. needs an explanation, not just an answer

Names the specific remaining step, substituting the input back into the original equation, that would produce the value. . Worth 1 point.

2. A garden that borrows one side from the house . Application, 10 points. Question 2 of 5.

A homeowner is building a rectangular vegetable garden against the back wall of the house, so the wall supplies one side and $72$ metres of fencing are available for the other three sides.

Part A.
Let $x$ be the length of each of the two sides that run out from the wall. Write the length of the side parallel to the wall in terms of $x$ , and write the garden's area $A$ as a quadratic in $x$ .

Model the situation Name your unknown first, then write every other quantity in terms of that one letter. 3 points
Part B.
Find the dimensions, the length of each side out from the wall and the length of the side parallel to it, that give the greatest possible area.

Carry your own answer forward Use the area expression you set up in part A, even if it is not fully expanded.

Solve and show your work Write each step out, and end with the value and its units. 4 points
Part C.
State the greatest possible area of the garden, and explain in one sentence why this is a different question from the one part B answered.

Carry your own answer forward Substitute your optimizing value of $x$ from part B into the area expression from part A, or multiply the two dimensions you found.

Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 3 points

Hints

One at a time, each one a step further than the last. Take only as many as you need.

Hint 1 of 3
This garden only needs fencing on three sides, since the wall supplies the fourth for free. Do not spend any of the $72$ metres fencing a side that is already there.
Hint 2 of 3 · Part A
Let one pair of equal sides be $x$ . Whatever fence is left after those two sides is what forms the side parallel to the wall; write that leftover length in terms of $x$ before multiplying for area.
Hint 3 of 3 · Part B
Once the area is a quadratic in $x$ , its vertex input comes from $x=-\tfrac{b}{2a}$ . Find that first, and only then get the parallel side by substituting it into its own expression, not into the area formula.

Answer and solution

Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Recognizes that the wall replaces one side, so only two $x$ -length sides and one parallel side need fencing, and gets the parallel side by subtracting those two sides from the total fence, not from a full four-sided perimeter. . Worth 2 points.

Multiplies the two dimensions to express the area as a single quadratic in $x$ . . Worth 1 point.

Part B 4 points

Finds the optimizing $x$ by applying $x=-\tfrac{b}{2a}$ to the area expression from part A. . Worth 2 points.

Substitutes that $x$ into the PARALLEL SIDE expression, not the area formula, to get the second dimension. . Worth 1 point.

Reports the two side lengths only, the dimensions asked for, not the area they enclose. . Worth 1 point.

Part C 3 points

Correctly computes the area from the two dimensions found in part B. . Worth 2 points.

Explains that the area is the vertex's optimal VALUE, while part B's dimensions are the optimizing INPUT, so the two are not interchangeable answers. . Worth 1 point. needs an explanation, not just an answer

3. What the sign of a decides, and does not . Reasoning, 10 points. Question 3 of 5.

A claim: the sign of a quadratic quantity's leading coefficient does not just decide between a maximum and a minimum, it also decides whether that best value itself is positive. Specifically, a negative leading coefficient always produces a positive maximum, and a positive leading coefficient always produces a positive minimum.

Part A.
Refute the first half of the claim: produce one quadratic quantity with a negative leading coefficient whose maximum value is not positive, and show the vertex computation that proves it.

Construct a counterexample Give one specific case, and show it breaks the claim. 4 points
Part B.
The claim conflates two separate facts about a vertex. State the ONE thing the sign of $a$ actually decides, and state what it does NOT decide about the resulting value.

Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 3 points
Part C.
Now test the second half of the claim: compute the actual minimum value of $y=x^2-2x-8$ , and state whether that number confirms or refutes the claim's promise about positive leading coefficients.

Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 3 points

Hints

One at a time, each one a step further than the last. Take only as many as you need.

Hint 1 of 3
The claim bundles two separate jobs into one: deciding up-versus-down, and deciding the sign of the resulting number. Test whether the sign of $a$ can really do both jobs at once.
Hint 2 of 3 · Part A
Pick any downward-opening quadratic ( $a<0$ ) whose constant term is a large enough negative number that the whole thing stays negative even at its peak. Compute its vertex to check.
Hint 3 of 3 · Part C
Run the exact same kind of check on an upward-opening quadratic. A minimum can sit below the horizontal axis just as easily as a maximum can, whatever the sign of $a$ was.

Answer and solution

Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.

Worked solution

Part A

The claim says a negative leading coefficient always produces a positive maximum. To disprove a claim about ALL such quadratics, one honest counterexample is enough.

Take $y=-2x^2-4x-9$ , where $a=-2<0$ . Find its vertex.

$x=-\frac{b}{2a}=-\frac{-4}{2(-2)}=\frac{4}{-4}=-1.$

Substitute back for the value.

$y=-2(-1)^2-4(-1)-9=-2+4-9=-7.$

The vertex is a maximum, since $a<0$ , but its value is $-7$ , a negative number. The claim promised a positive maximum; this one is not. That single example is enough to refute the claim as stated for every quadratic with $a<0$ .

Part B

Go back to the argument behind the vertex: writing $y=a(x-h)^2+k$ , the sign of $a$ decides only whether $y\ge k$ or $y\le k$ , that is, whether the vertex is a floor or a ceiling. Nothing in that argument touches the sign of $k$ itself.

$y=a(x-h)^2+k \;\Rightarrow\; \begin{cases}a>0: & y\ge k \\ a<0: & y\le k\end{cases}$

The value $k$ is set by completing the square on all of $a$ , $b$ , and $c$ together, not by $a$ alone. Two quadratics can share the same $a$ and still have their vertex sit above the axis, on it, or below it, depending on $b$ and $c$ . So the sign of $a$ answers whether the vertex opens up or down, never whether the resulting number is positive or negative.

Part C

The claim's second half concerns quadratics with $a>0$ . Test it on $y=x^2-2x-8$ , where $a=1>0$ .

$x=-\frac{b}{2a}=-\frac{-2}{2(1)}=1.$

Substitute back for the value.

$y=(1)^2-2(1)-8=1-2-8=-9.$

The vertex is a minimum, since $a>0$ , but its value is $-9$ , a negative number. The second half of the claim fails for exactly the same reason the first half did: the sign of $a$ never touches the sign of the extreme value.

In one line

A negative leading coefficient does not guarantee a positive maximum: $y=-2x^2-4x-9$ has $a=-2<0$ but its maximum value is $-7$ . The sign of $a$ decides only up versus down, never the sign of the resulting number, which also depends on $b$ and $c$ ; the same failure shows up on the positive side, since $y=x^2-2x-8$ has $a=1>0$ but a minimum of $-9$ .

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Chooses a specific quadratic with $a<0$ to test the claim, rather than arguing about quadratics in general. . Worth 1 point.

Correctly finds the vertex of the chosen quadratic, including substituting back for the value, and gets a negative result. . Worth 2 points.

States that this one example is enough to refute a claim asserted for every quadratic with $a<0$ . . Worth 1 point.

Part B 3 points

Separates what the sign of $a$ actually decides, maximum versus minimum, from what it does not decide, the sign of the resulting value. . Worth 2 points. needs an explanation, not just an answer

States that the value's sign depends on the quadratic's other coefficients too, not on $a$ in isolation. . Worth 1 point.

Part C 3 points

Correctly finds the vertex of $y=x^2-2x-8$ , including substituting back for the value. . Worth 2 points.

States plainly whether the computed value confirms or refutes the claim's second half. . Worth 1 point.

4. A price the parabola would not pick . Application, 13 points. Question 4 of 5.

A minor-league baseball team sells $650$ tickets to a home game at a price of $14$ dollars each. Ticket-office data shows that each $1$ dollar increase in price is expected to cost the team $25$ ticket sales.

Part A.
Let $x$ be the number of $1$ dollar price increases from $14$ dollars. Write the number of tickets sold, and the revenue $R$ , as expressions in $x$ .

Model the situation Name your unknown first, then write every other quantity in terms of that one letter. 3 points
Part B.
Find the ticket price that maximizes revenue, and find the maximum revenue itself. Report both numbers, and be clear about which is which.

Carry your own answer forward Use the revenue expression you built in part A, even if it is not fully expanded.

Solve and show your work Write each step out, and end with the value and its units. 6 points
Part C.
The team's manager instead sets the price at $24$ dollars. Compute the revenue at that price, and explain in one sentence why it comes out below the maximum from part B, tying your answer to the shape of the revenue parabola.

Carry your own answer forward Use the revenue expression from part A.

Compare the two methods Say what each one costs you, and when you would reach for it. 4 points

Hints

One at a time, each one a step further than the last. Take only as many as you need.

Hint 1 of 3
Two different variables are floating around here: $x$ , the number of $1$ dollar increases, and the actual ticket price, $14+x$ . Keep straight which one each part wants.
Hint 2 of 3 · Part B
Build the revenue expression first, then use $x=-\tfrac{b}{2a}$ on it. That hands you $x$ , not the price: add it to $14$ before you report a price.
Hint 3 of 3 · Part C
Revenue is a downward parabola with exactly one highest point. Any price other than the one at that peak sits somewhere on one of the two falling sides.

Answer and solution

Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.

Worked solution

Part A

Each $1$ dollar price increase costs $25$ ticket sales, so after $x$ increases the price is $14+x$ dollars and the number sold is $650-25x$ .

Revenue is price times quantity:

$R=(14+x)(650-25x).$

This is the model; expanding it is the next part's job.

Part B

Expand the model from part A to see its coefficients.

$R=(14+x)(650-25x)=9100+300x-25x^2=-25x^2+300x+9100.$

Since $a=-25<0$ , the vertex is a maximum. Find the optimizing $x$ .

$x=-\frac{b}{2a}=-\frac{300}{2(-25)}=-\frac{300}{-50}=6.$

The question asks for the PRICE, not $x$ itself, so add it to the base price: $14+6=20$ dollars.

Substitute $x=6$ back into $R$ for the maximum revenue.

$R=-25(6)^2+300(6)+9100=-900+1800+9100=10000.$

The maximum revenue is $10000$ dollars, a separate number from the $20$ dollar price.

Part C

A price of $24$ dollars is $10$ dollars above the base price of $14$ , so $x=10$ . Substitute into the revenue expression from part A.

$R=-25(10)^2+300(10)+9100=-2500+3000+9100=9600.$

The revenue at $24$ dollars is $9600$ dollars, below the $10000$ -dollar maximum from part B. Revenue is a downward parabola with exactly one highest point, at $x=6$ ; moving away from that single peak in either direction, including to $x=10$ , can only land on a lower point of the curve.

In one line

$R=(14+x)(650-25x)$ , where $x$ is the number of $1$ dollar increases; the revenue-maximizing price is $20$ dollars, giving a maximum revenue of $10000$ dollars; and the manager's $24$ dollar price gives only $9600$ dollars, since it falls away from the vertex of the downward revenue parabola.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Builds the number of tickets sold by taking away $25$ sales for every $1$ dollar increase from the starting count, and builds the price by adding that same number of increases to the starting price. . Worth 2 points.

Multiplies these two expressions to state the revenue $R$ as a single product. . Worth 1 point.

Part B 6 points

Finds the optimizing $x$ by applying $x=-\tfrac{b}{2a}$ to the expanded revenue expression. . Worth 2 points.

Converts $x$ into the actual ticket PRICE, $14+x$ , since price rather than $x$ is what the question asks for. . Worth 1 point.

Substitutes the optimizing $x$ back into $R$ to compute the maximum revenue. . Worth 2 points.

Reports the maximum revenue as a separate figure from the price, not the two run together. . Worth 1 point.

Part C 4 points

Correctly computes the revenue at the manager's proposed price. . Worth 2 points.

Explains that because the revenue parabola opens downward with a single peak, any price other than the one from part B necessarily gives a lower revenue. . Worth 2 points. needs an explanation, not just an answer

5. Two platforms, one launch . Reasoning, 10 points. Question 5 of 5.

A ball is launched straight up with an initial speed of $64$ feet per second. One version of the experiment launches it from a $10$ foot platform, so its height is $h=-16t^2+64t+10$ ; a second version launches the same ball from a $40$ foot platform, so its height is $h=-16t^2+64t+40$ . Everything else about the launch is the same.

Part A.
Find the time at which each version reaches its peak height.

Solve and show your work Write each step out, and end with the value and its units. 3 points
Part B.
Find the maximum height each version reaches.

Solve and show your work Write each step out, and end with the value and its units. 3 points
Part C.
Using your answers from parts A and B, compare the two peak times with each other, and separately compare the two peak heights with each other. Describe what you notice in each comparison, and explain it in terms of which of the coefficients $a$ , $b$ , $c$ the optimizing input depends on, and which the optimal value additionally depends on.

Carry your own answer forward Use whatever times you found in part A and whatever heights you found in part B, even if a number did not come out as expected.

Compare the two methods Say what each one costs you, and when you would reach for it. 4 points

Hints

One at a time, each one a step further than the last. Take only as many as you need.

Hint 1 of 3
The two versions of this launch share the same $-16t^2+64t$ part and differ only in the constant added at the end. Ask which of the vertex's two coordinates that constant can actually move.
Hint 2 of 3 · Part A
The peak TIME comes from $x=-\tfrac{b}{2a}$ , a formula built only out of $a$ and $b$ . Check what changes between the two platforms' equations and what does not.
Hint 3 of 3 · Part C
Once you have both peak times and both peak heights, line them up in a small table before you write the explanation. Patterns are easier to name once they are side by side.

Answer and solution

Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.

Worked solution

Part A

Both height equations share the same $-16t^2+64t$ part, so $a=-16$ and $b=64$ for each. The peak time depends only on those two coefficients.

$t=-\frac{b}{2a}=-\frac{64}{2(-16)}=-\frac{64}{-32}=2.$

That computation never used either constant term, so both the $10$ foot platform and the $40$ foot platform peak at the same time, $t=2$ seconds.

Part B

Substitute $t=2$ into each platform's own height equation, including its own starting height.

For the $10$ foot platform:

$h=-16(2)^2+64(2)+10=-64+128+10=74.$

For the $40$ foot platform:

$h=-16(2)^2+64(2)+40=-64+128+40=104.$

The two maximum heights are $74$ feet and $104$ feet.

Part C

Line the two comparisons up. The peak times from part A are the same number for both platforms, because $x=-\tfrac{b}{2a}$ is built only out of $a$ and $b$ , and those two coefficients never changed between the versions.

The peak heights from part B are not the same: substituting the shared time back into $h=-16t^2+64t+c$ adds in $c$ , the starting height, which is the one number that did change between the platforms, by $40-10=30$ feet. That is exactly the gap between the two heights found in part B.

$\underbrace{104-74}_{\text{height gap}}=30=\underbrace{40-10}_{\text{platform gap}}.$

So the optimizing input answers a question that $c$ has no say in, while the optimal value is a question $c$ always gets a vote on.

In one line

Both platforms peak at $t=2$ seconds, since the peak time $x=-\tfrac{b}{2a}$ only uses $a$ and $b$ , which are identical for the two versions. Their peak heights differ, by exactly the $30$ foot gap between the platforms, because substituting back in also brings in $c$ , the one coefficient that changed.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Correctly computes the peak TIME for each of the two platforms using $x=-\tfrac{b}{2a}$ . . Worth 2 points.

Reports a time for each platform, not a height. . Worth 1 point.

Part B 3 points

Substitutes each platform's own peak time into its OWN height equation to find each maximum height. . Worth 2 points.

Reports a height for each platform, distinct from the times found in part A. . Worth 1 point.

Part C 4 points

Accurately states how the two peak times compare to each other, and separately how the two peak heights compare to each other, based on the student's own computed values. . Worth 2 points.

Explains the pattern by identifying that the optimizing input depends only on $a$ and $b$ , while the optimal value additionally depends on $c$ . . Worth 2 points. needs an explanation, not just an answer

1. Two vertices, two different questions . Foundational, 9 points. Question 1 of 5.

2. A garden that borrows one side from the house . Application, 10 points. Question 2 of 5.

3. What the sign of a decides, and does not . Reasoning, 10 points. Question 3 of 5.

4. A price the parabola would not pick . Application, 13 points. Question 4 of 5.

5. Two platforms, one launch . Reasoning, 10 points. Question 5 of 5.

Marking rubrics

1. Two vertices, two different questions 9 points

2. A garden that borrows one side from the house 10 points

3. What the sign of a decides, and does not 10 points

4. A price the parabola would not pick 13 points

5. Two platforms, one launch 10 points