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Graphing Quadratics and Inequalities: Chapter Test

20 multiple-choice questions and 10 free-response questions, drawn from across the chapter and mixed together.

Free response

10 questions in parts, 103 points in total. Work them out on paper. There are no hints here: reveal each question's answer, worked solution, and rubric when you are ready to mark that one.

Free response · work it on paper

Question 1 of 10

1. What a parabola tells you before you plot it . 8 points. Question 1 of 10.

The quadratic $y = 2x^2 - 8x + 3$ is written in standard form. Several features of its graph can be read straight from the three coefficients, with no plotting and no solving.
1. Part A.
  State whether the graph opens upward or downward and whether it is narrower or wider than $y = x^2$ , and give the coordinates of its $y$ -intercept.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 3 points
2. Part B.
  Find the axis of symmetry of $y = 2x^2 - 8x + 3$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 2 points
3. Part C.
  Say which coefficient each of your answers in parts A and B came from, and name the one feature that the coefficient $b$ controls but $a$ and $c$ do not.
  
  Carry your own answer forward Refer to the answers you gave in parts A and B, whatever they were, and identify the coefficient standing behind each one.
  
  Explain why it works A sentence or two. Reasons, not steps. 3 points
The answer

Part A

It opens upward and is narrower than $y = x^2$ , and its $y$ -intercept is $(0, 3)$ .

Part B

$x = 2$ .

Part C

Direction and width came from $a$ , and the $y$ -intercept from $c$ ; neither used $b$ . The axis of symmetry came from $a$ and $b$ together. What $b$ controls, and $a$ and $c$ do not, is the horizontal position of the axis of symmetry, and with it the vertex.

Worked solution

Part A

Direction and width read from $a = 2$ : it is positive, so the graph opens upward, and $|a| = 2 > 1$ , so it is narrower than $y = x^2$ . The $y$ -intercept is the point where $x = 0$ , which leaves the constant term.

$a = 2 > 0 \Rightarrow \text{upward}, \quad |a| = 2 > 1 \Rightarrow \text{narrower}, \quad y(0) = 3 \Rightarrow (0, 3).$

Part B

The axis of symmetry is $x = -\tfrac{b}{2a}$ with $a = 2$ and $b = -8$ .

$x = -\frac{-8}{2(2)} = \frac{8}{4} = 2.$

Part C

Part A used only $a = 2$ for direction and width, and only $c = 3$ for the $y$ -intercept; the value of $b$ never entered either. Part B used both $a$ and $b$ in $x = -\tfrac{b}{2a}$ .

$\text{direction, width} \leftarrow a, \quad y\text{-intercept} \leftarrow c, \quad \text{axis} \leftarrow a, b.$

So $b$ is the only coefficient that moves the axis of symmetry; changing it slides the curve horizontally while its shape (from $a$ ) and its $y$ -intercept (from $c$ ) stay put.

In one line

$y = 2x^2 - 8x + 3$ opens upward and is narrower than $y = x^2$ (from $a = 2$ ), has $y$ -intercept $(0, 3)$ (from $c = 3$ ), and axis of symmetry $x = 2$ (from $a$ and $b$ ). Changing $b$ , with $a$ and $c$ fixed, moves the axis of symmetry and the vertex, which is why direction, width, and the $y$ -intercept do not depend on it.

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

Reads the direction and the width from the leading coefficient $a$ alone, not from $b$ or $c$ . . Worth 2 points.

Gives the $y$ -intercept as the ordered pair $(0, c)$ , not just the constant. . Worth 1 point.

Part B 2 points

Applies $x = -\tfrac{b}{2a}$ with the leading minus sign, not $x = \tfrac{b}{2a}$ . . Worth 1 point.

Simplifies the fraction to the correct axis value. . Worth 1 point.

Part C 3 points

Correctly attributes each earlier answer to the coefficient that produced it, tying direction and width to $a$ and the $y$ -intercept to $c$ . . Worth 2 points. needs an explanation, not just an answer

Names the horizontal position (the axis of symmetry) as what $b$ controls that $a$ and $c$ do not. . Worth 1 point.
2. A vertex is a point, not a number . 8 points. Question 2 of 10.

Two parabolas are on the table: $y = x^2 + 8x + 11$ and $y = -3(x - 2)^2 + 5$ . Both have a vertex, and each part asks about one of them.
1. Part A.
  Find the vertex of $y = x^2 + 8x + 11$ as a point.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Read the vertex of $y = -3(x - 2)^2 + 5$ directly from vertex form.
  
  Solve and show your work Write each step out, and end with the value and its units. 2 points
3. Part C.
  For $y = x^2 + 8x + 11$ , explain why the axis-of-symmetry value you used in part A is not, by itself, the vertex, and name the single step that turns it into the full point.
  
  Carry your own answer forward Argue from the $x$ -coordinate you found in part A, whatever it was, and the step that produces the second coordinate.
  
  Explain why it works A sentence or two. Reasons, not steps. 3 points
The answer

Part A

$(-4, -5)$ .

Part B

$(2, 5)$ .

Part C

That value is only the input, the first of the vertex's two coordinates. A vertex is a point $(x, y)$ , so it is incomplete until the height is found, which is done by substituting that input back into the equation.

Worked solution

Part A

The axis of symmetry gives the $x$ -coordinate, and substituting it back gives the height.

$x = -\frac{8}{2(1)} = -4, \qquad y = (-4)^2 + 8(-4) + 11 = 16 - 32 + 11 = -5.$

Part B

Vertex form $y = a(x - h)^2 + k$ has vertex $(h, k)$ . Here $x - 2$ gives $h = 2$ and the loose constant gives $k = 5$ .

$(h, k) = (2, 5).$

Part C

The number $x = -4$ locates only WHERE the turning point sits along the $x$ -axis. A vertex is a point with two coordinates, and $x = -\tfrac{b}{2a}$ produces just the first.

$\text{input } x = -4 \;\longrightarrow\; \text{substitute back} \;\longrightarrow\; y = -5.$

The one remaining step is to substitute that input into the original equation to get the height, giving the full point $(-4, -5)$ . The input alone is not the vertex.

In one line

$y = x^2 + 8x + 11$ has vertex $(-4, -5)$ , and $y = -3(x - 2)^2 + 5$ has vertex $(2, 5)$ . The axis value $x = -4$ is only the input, the first coordinate; the vertex is complete only after substituting it back to get the height $-5$ .

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

Finds the $x$ -coordinate from $-\tfrac{b}{2a}$ and substitutes it back to get the height. . Worth 2 points.

Reports the vertex as an ordered pair, both coordinates present. . Worth 1 point.

Part B 2 points

Matches the expression against $y = a(x - h)^2 + k$ rather than expanding it. . Worth 1 point.

Reports the vertex as an ordered pair read straight from the form. . Worth 1 point.

Part C 3 points

Explains that a vertex is a point with two coordinates, so the axis value is only the input, not the whole vertex. . Worth 2 points. needs an explanation, not just an answer

Names substituting the input back into the equation as the step that produces the height. . Worth 1 point.
3. A circle pinned down by a diameter . 11 points. Question 3 of 10.

A circle has a diameter whose endpoints are $P(-1, 2)$ and $Q(5, 10)$ .
1. Part A.
  Find the center of the circle.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Write the standard equation of the circle, using the fact that $Q$ lies on it.
  
  Carry your own answer forward Use the center you found in part A as $(h, k)$ , and get $r^2$ as the squared distance from that center out to an endpoint.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
3. Part C.
  Determine whether the point $(5, 3)$ lies inside, on, or outside this circle.
  
  Carry your own answer forward Use your own center and $r^2$ from parts A and B; compare the point's squared distance from the center with $r^2$ .
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 4 points
The answer

Part A

$(2, 6)$ .

Part B

$(x - 2)^2 + (y - 6)^2 = 25$ .

Part C

Inside: the point's squared distance from the center is $18$ , less than $r^2 = 25$ .

Worked solution

Part A

The center of a circle is the midpoint of any diameter, found by averaging the endpoints coordinate by coordinate.

$\left(\frac{-1 + 5}{2}, \frac{2 + 10}{2}\right) = (2, 6).$

Part B

Only $r^2$ is needed, so square the distance from the center to $Q$ without taking a root.

$r^2 = (5 - 2)^2 + (10 - 6)^2 = 9 + 16 = 25.$

Substitute the center and $r^2$ into $(x - h)^2 + (y - k)^2 = r^2$ .

$(x - 2)^2 + (y - 6)^2 = 25.$

Part C

Compare the squared distance from the center to $(5, 3)$ against $r^2$ , no root needed.

$(5 - 2)^2 + (3 - 6)^2 = 9 + 9 = 18 < 25.$

The squared distance is less than $r^2$ , so the point lies inside the circle.

In one line

The center is the midpoint $(2, 6)$ , and $r^2 = 25$ , so the circle is $(x - 2)^2 + (y - 6)^2 = 25$ ; the point $(5, 3)$ lies inside it, since its squared distance from the center, $18$ , is less than $r^2 = 25$ .

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 the midpoint of the diameter, averaging each coordinate separately. . Worth 2 points.

Reports the center as a single ordered pair. . Worth 1 point.

Part B 4 points

Computes $r^2$ as the squared distance from the center to an endpoint, skipping the unnecessary root. . Worth 2 points.

Writes the standard equation with the correct center signs and $r^2$ . . Worth 2 points.

Part C 4 points

Computes the squared distance from the center to the point correctly. . Worth 2 points.

Compares that squared distance with $r^2$ (not $r$ ) to reach the inside, on, or outside verdict. . Worth 2 points.
4. Classifying two completed-square equations . 10 points. Question 4 of 10.

Two equations are given: $x^2 + y^2 - 10x + 4y + 29 = 0$ and $x^2 + y^2 - 10x + 4y + 33 = 0$ . They differ only in their constant term.
1. Part A.
  Rewrite the first equation in the form $(x - h)^2 + (y - k)^2 = R$ by grouping each variable into a perfect square. State $h$ , $k$ , and $R$ , and say precisely what the graph is.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Rewrite the second equation in the same form, $(x - h)^2 + (y - k)^2 = R$ , the same way. State $h$ , $k$ , and $R$ , and say precisely what the graph is.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  Using the fact that a sum of two real squares is never negative, explain what each sign of $R$ (positive, zero, negative) means for the graph of $(x - h)^2 + (y - k)^2 = R$ , and check that against your two results.
  
  Carry your own answer forward Compare the two values of $R$ you found in parts A and B, whatever they came out to be, against the three cases.
  
  Explain why it works A sentence or two. Reasons, not steps. 4 points
The answer

Part A

$(x - 5)^2 + (y + 2)^2 = 0$ , so $h = 5$ , $k = -2$ , $R = 0$ ; the graph is the single point $(5, -2)$ .

Part B

$(x - 5)^2 + (y + 2)^2 = -4$ , so $h = 5$ , $k = -2$ , $R = -4$ ; since $R < 0$ the graph is empty (no real points).

Part C

The left side is never negative. If $R > 0$ it traces a circle of radius $\sqrt{R}$ ; if $R = 0$ both squares must be $0$ , giving one point; if $R < 0$ nothing satisfies it, so the graph is empty. Parts A and B are the $R = 0$ and $R < 0$ cases; a genuine circle needs $R > 0$ .

Worked solution

Part A

Group each variable and move the constant across.

$(x^2 - 10x) + (y^2 + 4y) = -29 \;\Rightarrow\; (x - 5)^2 + (y + 2)^2 = -29 + 25 + 4 = 0.$

A sum of two real squares equals $0$ only when each square is $0$ , forcing $x = 5$ and $y = -2$ . The graph is the single point $(5, -2)$ .

Part B

The grouping is identical; only the moved constant differs.

$(x^2 - 10x) + (y^2 + 4y) = -33 \;\Rightarrow\; (x - 5)^2 + (y + 2)^2 = -33 + 25 + 4 = -4.$

A sum of two real squares can never be negative, so no real point satisfies the equation and the graph is empty.

Part C

For real $x$ and $y$ , each of $(x - h)^2$ and $(y - k)^2$ is at least $0$ , so their sum is at least $0$ .

$(x - h)^2 + (y - k)^2 \ge 0.$

If $R > 0$ , the sum can equal it in many ways, tracing a circle of radius $\sqrt{R}$ . If $R = 0$ , the only way is both squares $0$ , a single point. If $R < 0$ , no real point works, so the graph is empty. Part A gave $R = 0$ (a point) and part B gave $R < 0$ (empty); only $R > 0$ is a genuine circle.

In one line

Completing the square on the first equation gives $(x - 5)^2 + (y + 2)^2 = 0$ , the single point $(5, -2)$ ; the second gives $(x - 5)^2 + (y + 2)^2 = -4$ , an empty graph. Since a sum of two real squares is never negative, $R > 0$ gives a circle, $R = 0$ a point, and $R < 0$ nothing.

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

Completes the square on each variable, adding both constants to the right side correctly. . Worth 2 points.

States $R$ and names what the graph is, rather than leaving it as an equation. . Worth 1 point.

Part B 3 points

Carries out the same completing-the-square step and reaches the value of $R$ correctly. . Worth 2 points.

Reads what the graph is from the sign of $R$ obtained, classifying it correctly. . Worth 1 point.

Part C 4 points

Argues from a sum of two real squares being nonnegative to what each sign of $R$ produces, matching the two results from parts A and B. . Worth 3 points. needs an explanation, not just an answer

States that only $R > 0$ gives a genuine circle, naming the point and empty cases correctly. . Worth 1 point.
5. One inequality, three ways to say its answer . 11 points. Question 5 of 10.

Consider the inequality $x^2 - 2x - 8 \le 0$ .
1. Part A.
  The inequality already has $0$ on the right. Factor the quadratic and state its two roots.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
2. Part B.
  Build the sign chart from the roots and state the solution of $x^2 - 2x - 8 \le 0$ as an inequality.
  
  Carry your own answer forward Build the sign chart from the two roots you found in part A, whatever they turned out to be.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Write the same solution in interval notation, and describe its number-line picture: which circles are open or closed, and where the shading runs.
  
  Carry your own answer forward Describe the same solution you reported in part B, using its own boundary numbers, whatever they were.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 4 points
The answer

Part A

$x^2 - 2x - 8 = (x - 4)(x + 2)$ ; roots $x = -2$ and $x = 4$ .

Part B

$-2 \le x \le 4$ .

Part C

$[-2, 4]$ ; closed circles at both roots (the symbol is inclusive), with the segment between them shaded and everything outside left unshaded.

Worked solution

Part A

Two numbers multiplying to $-8$ and adding to $-2$ are $-4$ and $2$ .

$x^2 - 2x - 8 = (x - 4)(x + 2),$

so the roots are $x = 4$ and $x = -2$ .

Part B

The roots split the line into three intervals; test one point in each. The parabola opens upward ( $a = 1 > 0$ ), so it is negative between the roots and positive outside.

$x = 0: (0 - 4)(0 + 2) = -8 < 0.$

The symbol $\le$ is inclusive, so the roots are kept: $-2 \le x \le 4$ .

Part C

The solution is a single closed segment, so interval notation uses square brackets.

$[-2, 4].$

The inclusive $\le$ puts CLOSED circles at both $-2$ and $4$ , and the shading runs across the interval between them, where the quadratic is negative, leaving the two outside rays unshaded.

In one line

$x^2 - 2x - 8 = (x - 4)(x + 2)$ with roots $-2$ and $4$ ; the upward parabola is negative between them, so the solution of $\le 0$ is $-2 \le x \le 4$ , which is $[-2, 4]$ , with closed circles at both roots and the segment between them shaded.

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

Factors the quadratic correctly. . Worth 2 points.

Reads both roots off the factors. . Worth 1 point.

Part B 4 points

Reads the sign of the upward parabola on each interval, selecting where it is negative. . Worth 2 points.

Chooses the interval matching the sign the inequality asks for, with endpoints matching the symbol. . Worth 2 points.

Part C 4 points

Writes the solution in interval notation with square brackets to match the inclusive symbol. . Worth 2 points.

Describes the number-line picture correctly, with the circles and shading matching the symbol and the chosen interval. . Worth 2 points.
6. Two crossings is not a law . 10 points. Question 6 of 10.

A claim: 'the graph of every quadratic $y = ax^2 + bx + c$ crosses the $x$ -axis at exactly two points.' So many textbook parabolas cross twice that the claim sounds safe.
1. Part A.
  Refute the claim: give one specific quadratic for which it fails, and confirm the failure with its discriminant.
  
  Construct a counterexample Give one specific case, and show it breaks the claim. 3 points
2. Part B.
  For $y = x^2 - 6x + 9$ , state how many $x$ -intercepts the graph has and where the vertex sits relative to the $x$ -axis.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 3 points
3. Part C.
  State the three-way relationship between the sign of the discriminant and the number of $x$ -intercepts, and explain why restricting the claim to upward-opening parabolas still cannot rescue 'always two.'
  
  Justify your claim State the claim, then give the reason it has to be true. 4 points
The answer

Part A

$y = x^2 + 2x + 5$ has $b^2 - 4ac = 4 - 20 = -16 < 0$ , so it has no $x$ -intercepts, not two. One counterexample refutes a claim about every parabola.

Part B

One $x$ -intercept. The discriminant is $36 - 36 = 0$ , so the vertex sits exactly on the $x$ -axis.

Part C

$D > 0$ gives two intercepts, $D = 0$ one, $D < 0$ none, whatever the sign of $a$ . Restricting to upward parabolas cannot force $D > 0$ : the part A example already has $a = 1 > 0$ and no intercepts, its vertex resting above the axis.

Worked solution

Part A

One honest counterexample breaks a claim about EVERY parabola. Take $y = x^2 + 2x + 5$ .

$b^2 - 4ac = (2)^2 - 4(1)(5) = 4 - 20 = -16 < 0.$

A negative discriminant means no real roots, so this parabola has zero $x$ -intercepts, not two.

Part B

Compute the discriminant with $a = 1$ , $b = -6$ , $c = 9$ .

$(-6)^2 - 4(1)(9) = 36 - 36 = 0.$

A zero discriminant means exactly one real root, which happens geometrically when the vertex rests on the $x$ -axis rather than above or below it.

Part C

The roots come from the quadratic formula,

$x = \frac{-b \pm \sqrt{D}}{2a},$

so a positive $D$ gives two real roots, $D = 0$ gives one, and $D < 0$ gives none.

Restricting to $a > 0$ does not help: the sign of $D$ is decided by $a$ , $b$ , and $c$ together, and opening upward forces none of them. The part A quadratic has $a = 1 > 0$ yet $D = -16 < 0$ , so an upward parabola with no crossings already exists.

In one line

The claim fails: $y = x^2 + 2x + 5$ has $D = -16 < 0$ and no $x$ -intercepts, while $y = x^2 - 6x + 9$ has $D = 0$ and exactly one, its vertex on the axis. In general $D > 0$ gives two, $D = 0$ gives one, and $D < 0$ gives none, and the sign of $a$ alone never forces $D > 0$ .

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

Chooses a specific quadratic and computes its discriminant correctly. . Worth 2 points.

States that the count found contradicts the claim of exactly two. . Worth 1 point.

Part B 3 points

Computes the discriminant and reads the intercept count from it. . Worth 2 points.

Describes the vertex's position relative to the $x$ -axis, consistent with the count it found. . Worth 1 point.

Part C 4 points

States the three-way discriminant classification completely. . Worth 2 points.

Explains, with a case where $a > 0$ but the count is not two, why restricting to upward parabolas fails. . Worth 2 points. needs an explanation, not just an answer
7. A pen with one side already built . 10 points. Question 7 of 10.

A rectangular pen is built against a straight barn wall, so the wall forms one side and $60$ metres of fencing are used for the other three sides.
1. 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 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
2. Part B.
  Find the dimensions, each side out from the wall and the side parallel to it, that give the greatest area.
  
  Carry your own answer forward Use the area expression you set up in part A, even if it is not expanded.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  State the greatest possible area, and explain in one sentence why this is a different question from the one part B answered.
  
  Carry your own answer forward Substitute the two side lengths you found in part B into your area expression from part A.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 3 points
The answer

Part A

The parallel side is $60 - 2x$ metres, so the area is $A = x(60 - 2x) = 60x - 2x^2$ square metres.

Part B

Each side out from the wall is $15$ metres, and the side parallel to the wall is $30$ metres.

Part C

The greatest area is $450$ square metres. That is the optimal value, the vertex height; the $15$ and $30$ metres from part B are the optimizing input, the dimensions. They answer two different questions about the same vertex.

Worked solution

Part A

The wall supplies one side free, so the $60$ metres cover only two sides of length $x$ and the one parallel side. What is left after the two $x$ -sides forms that parallel side.

$\text{parallel side} = 60 - 2x, \qquad A = x(60 - 2x) = 60x - 2x^2.$

Part B

Since $a = -2 < 0$ , the area opens downward and its vertex is the maximum. Find the optimizing input.

$x = -\frac{60}{2(-2)} = 15.$

Substitute into the parallel-side expression, not the area, for the second dimension: $60 - 2(15) = 30$ .

Part C

Multiply the two dimensions from part B to get the value the vertex represents.

$A = 15 \times 30 = 450.$

The dimensions are the optimizing input, WHERE the maximum occurs; the area $450$ is the optimal value, the height the vertex reaches. Reporting $15$ and $30$ again would re-answer part B, not this part.

In one line

The area is $A = x(60 - 2x)$ with parallel side $60 - 2x$ ; the greatest area comes from $15$ metres out from the wall and $30$ metres parallel to it, and the greatest area itself is $450$ square metres, a separate number from those two dimensions.

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

Gets the parallel side by subtracting the two $x$ -sides from the total fence, not from a full four-sided perimeter. . Worth 2 points.

Multiplies the two dimensions to write 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. . Worth 2 points.

Substitutes that $x$ into the parallel-side expression, not the area formula, for the second dimension. . Worth 1 point.

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

Part C 3 points

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

Explains that the area is the optimal value while the dimensions are the optimizing input, so the two are not interchangeable. . Worth 1 point. needs an explanation, not just an answer
8. A price the parabola prefers . 13 points. Question 8 of 10.

A theater sells $300$ tickets to a show at $20$ dollars each. Market research shows that each $1$ dollar drop in the price sells $30$ more tickets.
1. Part A.
  Let $x$ be the number of $1$ dollar price drops from $20$ dollars. Write the ticket price, the number of tickets sold, and the revenue $R$ , each in terms of $x$ .
  
  Model the situation Name your unknown first, then write every other quantity in terms of that one letter. 3 points
2. Part B.
  Find the ticket price that maximizes revenue, and the maximum revenue itself. Report both, and be clear which is which.
  
  Carry your own answer forward Use the revenue expression you built in part A, expanding it if needed.
  
  Solve and show your work Write each step out, and end with the value and its units. 6 points
3. Part C.
  The manager instead charges $18$ dollars. Compute the revenue at that price, and explain in one sentence why it comes out below the maximum, 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
The answer

Part A

The price is $20 - x$ dollars, the number sold is $300 + 30x$ , and the revenue is $R = (20 - x)(300 + 30x)$ dollars.

Part B

The revenue-maximizing price is $15$ dollars (at $x = 5$ ), and the maximum revenue is $6750$ dollars, a separate figure from the price.

Part C

At a price of $18$ dollars the revenue is $6480$ dollars, below the $6750$ -dollar maximum. Revenue is a downward parabola with a single peak, so any price other than the optimal one lands on a falling side.

Worked solution

Part A

Each $1$ dollar drop lowers the price by $1$ and adds $30$ sales, so after $x$ drops the price is $20 - x$ and the number sold is $300 + 30x$ . Revenue is price times quantity.

$R = (20 - x)(300 + 30x).$

Part B

Expand to read the coefficients: $R = -30x^2 + 300x + 6000$ , which opens downward. Find the optimizing input.

$x = -\frac{300}{2(-30)} = 5.$

The price asked for is $20 - x = 15$ dollars. Substitute $x = 5$ back for the revenue: $R = -30(25) + 300(5) + 6000 = 6750$ .

Part C

A price of $18$ dollars is $2$ dollars below $20$ , so $x = 2$ .

$R = -30(2)^2 + 300(2) + 6000 = -120 + 600 + 6000 = 6480.$

This is below the $6750$ maximum. Revenue is a downward parabola with exactly one highest point at $x = 5$ , so moving away from it in either direction can only lower the revenue.

In one line

$R = (20 - x)(300 + 30x)$ , where $x$ is the number of $1$ dollar drops; the revenue-maximizing price is $15$ dollars, giving a maximum revenue of $6750$ dollars, and the manager's $18$ dollar price gives only $6480$ dollars, since it sits 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 price by subtracting the drops and the quantity by adding $30$ per drop. . Worth 2 points.

Multiplies the two to state the revenue 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. . Worth 2 points.

Converts $x$ into the actual price $20 - x$ , since price rather than $x$ is asked for. . Worth 1 point.

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

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

Part C 4 points

Computes the revenue at the manager's price correctly. . Worth 2 points.

Explains that a downward parabola with a single peak makes any other price give a lower revenue. . Worth 2 points. needs an explanation, not just an answer
9. The one constant that keeps a parabola above the axis . 11 points. Question 9 of 10.

This question asks for every value of the number $k$ that makes $x^2 - 6x + k > 0$ hold for every real $x$ . The leading coefficient is positive, so the parabola opens upward.
1. Part A.
  An upward-opening quadratic stays above the axis for every real $x$ exactly when it has no real roots. Write the discriminant of $x^2 - 6x + k$ in terms of $k$ , and the condition on it that forces no real roots.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
2. Part B.
  Solve that condition for $k$ .
  
  Carry your own answer forward Solve whichever inequality in $k$ you derived in part A, whatever form it took.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Substitute the boundary value of $k$ into the original quadratic and simplify it. Use what you get to explain why that boundary value is excluded from your answer.
  
  Carry your own answer forward Substitute the boundary value of $k$ from part B, the value where your inequality stops being strict, into the original quadratic.
  
  Justify your claim State the claim, then give the reason it has to be true. 4 points
The answer

Part A

$b^2 - 4ac = 36 - 4k$ ; no real roots requires $36 - 4k < 0$ .

Part B

$k > 9$ .

Part C

At $k = 9$ the quadratic is $x^2 - 6x + 9 = (x - 3)^2$ , which equals $0$ at $x = 3$ instead of staying strictly positive there. So $k = 9$ fails the strict requirement, which is why the answer is $k > 9$ , not $k \ge 9$ .

Worked solution

Part A

With $a = 1$ , $b = -6$ , $c = k$ , the discriminant is

$b^2 - 4ac = (-6)^2 - 4(1)(k) = 36 - 4k.$

No real roots means a negative discriminant, so the condition is $36 - 4k < 0$ .

Part B

Solve the linear inequality in $k$ .

$36 - 4k < 0 \;\Rightarrow\; 36 < 4k \;\Rightarrow\; k > 9.$

Part C

At the boundary $k = 9$ , substitute into the original quadratic.

$x^2 - 6x + 9 = (x - 3)^2.$

This perfect square is $0$ at $x = 3$ , not greater than $0$ . The requirement was that the quadratic be strictly above the axis for EVERY $x$ , and at $x = 3$ it sits on the axis instead. So $k = 9$ is excluded, which is exactly why the strict inequality $k > 9$ leaves it out.

In one line

The discriminant of $x^2 - 6x + k$ is $36 - 4k$ , and requiring no real roots gives $36 - 4k < 0$ , so $k > 9$ . At the boundary $k = 9$ the quadratic is the perfect square $(x - 3)^2$ , which touches $0$ at $x = 3$ rather than staying strictly positive, so $k = 9$ is excluded and the answer is strict.

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

Writes the discriminant correctly in terms of $k$ . . Worth 2 points.

States the no-real-roots condition as the discriminant being negative. . Worth 1 point.

Part B 4 points

Solves the inequality for $k$ , handling the direction correctly. . Worth 2 points.

Reports the solution in $k$ with the boundary strictness handled correctly. . Worth 2 points.

Part C 4 points

Substitutes the boundary value into the original quadratic and simplifies it correctly. . Worth 2 points.

Explains why the boundary value fails the strict requirement, justifying its exclusion. . Worth 2 points. needs an explanation, not just an answer
10. How high, and whether it clears the mark . 11 points. Question 10 of 10.

A ball is thrown straight up; its height in feet after $t$ seconds is $h = -16t^2 + 48t + 40$ .
1. Part A.
  Find the time at which the ball reaches its peak, and the height it reaches there.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Determine whether the ball ever reaches a height of $80$ feet, using the discriminant of the equation $h = 80$ .
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 4 points
3. Part C.
  Explain how the peak height from part A already predicts the discriminant sign you found in part B, and say which of the coefficients $a$ , $b$ , $c$ the peak time depends on and which the peak height additionally depends on.
  
  Carry your own answer forward Use the peak height you found in part A and the result you found in part B.
  
  Compare the two methods Say what each one costs you, and when you would reach for it. 4 points
The answer

Part A

The peak is at $t = 1.5$ seconds (the optimizing input), and the peak height is $76$ feet (the optimal value).

Part B

No. Setting $h = 80$ gives $2t^2 - 6t + 5 = 0$ , whose discriminant is $36 - 40 = -4 < 0$ , so no real time reaches $80$ feet.

Part C

The peak height $76$ feet is the highest the ball reaches, so $80$ feet is impossible, exactly what the negative discriminant shows. The peak time used only $a$ and $b$ ; the peak height also used $c$ , the launch height, which is what put $80$ out of reach.

Worked solution

Part A

The peak time comes from $-\tfrac{b}{2a}$ , and the peak height from substituting it back.

$t = -\frac{48}{2(-16)} = 1.5, \qquad h = -16(1.5)^2 + 48(1.5) + 40 = -36 + 72 + 40 = 76.$

Part B

Set the height equal to $80$ and gather to one side.

$-16t^2 + 48t + 40 = 80 \;\Rightarrow\; 2t^2 - 6t + 5 = 0.$

The discriminant is $(-6)^2 - 4(2)(5) = 36 - 40 = -4 < 0$ , so there is no real solution: the ball never reaches $80$ feet.

Part C

The peak height is the greatest value $h$ ever takes, so a target above it can never be met.

$76 < 80 \;\Rightarrow\; \text{no real time gives a height of } 80 \text{ feet},$

which is exactly why part B's discriminant came out negative.

The peak time came from $-\tfrac{b}{2a}$ , using only $a$ and $b$ . The peak height came from substituting that time back, which brings in $c$ as well. So $c$ , the launch height, is the coefficient that decides how high the peak is, and therefore whether $80$ feet is reachable.

In one line

The ball peaks at $t = 1.5$ seconds (the input) at a height of $76$ feet (the value). It never reaches $80$ feet: $h = 80$ gives $2t^2 - 6t + 5 = 0$ with discriminant $-4 < 0$ . Since the peak height $76$ is already below $80$ , the negative discriminant is no surprise, and the peak height additionally depends on $c$ , the launch height, while the peak time uses only $a$ and $b$ .

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

Finds the peak time from $-\tfrac{b}{2a}$ . . Worth 2 points.

Substitutes back for the peak height and labels which value is the input and which the height. . Worth 1 point.

Part B 4 points

Forms the equation $h = 80$ , simplifies it, and computes its discriminant. . Worth 2 points.

Reads the verdict from the sign of the discriminant. . Worth 2 points.

Part C 4 points

Connects the peak height to the discriminant sign found in part B. . Worth 2 points.

Explains that the peak time depends only on $a$ and $b$ while the peak height additionally depends on $c$ . . Worth 2 points. needs an explanation, not just an answer

Marking rubrics

Work every question first. Then mark your own paper against these, giving yourself the points for each element you actually wrote down.

1. What a parabola tells you before you plot it 8 points

Part A 3

Reads the direction and the width from the leading coefficient $a$ alone, not from $b$ or $c$ . 2
Gives the $y$ -intercept as the ordered pair $(0, c)$ , not just the constant. 1

Part B 2

Applies $x = -\tfrac{b}{2a}$ with the leading minus sign, not $x = \tfrac{b}{2a}$ . 1
Simplifies the fraction to the correct axis value. 1

Part C 3

Correctly attributes each earlier answer to the coefficient that produced it, tying direction and width to $a$ and the $y$ -intercept to $c$ . 2
Names the horizontal position (the axis of symmetry) as what $b$ controls that $a$ and $c$ do not. 1

2. A vertex is a point, not a number 8 points

Part A 3

Finds the $x$ -coordinate from $-\tfrac{b}{2a}$ and substitutes it back to get the height. 2
Reports the vertex as an ordered pair, both coordinates present. 1

Part B 2

Matches the expression against $y = a(x - h)^2 + k$ rather than expanding it. 1
Reports the vertex as an ordered pair read straight from the form. 1

Part C 3

Explains that a vertex is a point with two coordinates, so the axis value is only the input, not the whole vertex. 2
Names substituting the input back into the equation as the step that produces the height. 1

3. A circle pinned down by a diameter 11 points

Part A 3

Uses the midpoint of the diameter, averaging each coordinate separately. 2
Reports the center as a single ordered pair. 1

Part B 4

Computes $r^2$ as the squared distance from the center to an endpoint, skipping the unnecessary root. 2
Writes the standard equation with the correct center signs and $r^2$ . 2

Part C 4

Computes the squared distance from the center to the point correctly. 2
Compares that squared distance with $r^2$ (not $r$ ) to reach the inside, on, or outside verdict. 2

4. Classifying two completed-square equations 10 points

Part A 3

Completes the square on each variable, adding both constants to the right side correctly. 2
States $R$ and names what the graph is, rather than leaving it as an equation. 1

Part B 3

Carries out the same completing-the-square step and reaches the value of $R$ correctly. 2
Reads what the graph is from the sign of $R$ obtained, classifying it correctly. 1

Part C 4

Argues from a sum of two real squares being nonnegative to what each sign of $R$ produces, matching the two results from parts A and B. 3
States that only $R > 0$ gives a genuine circle, naming the point and empty cases correctly. 1

5. One inequality, three ways to say its answer 11 points

Part A 3

Factors the quadratic correctly. 2
Reads both roots off the factors. 1

Part B 4

Reads the sign of the upward parabola on each interval, selecting where it is negative. 2
Chooses the interval matching the sign the inequality asks for, with endpoints matching the symbol. 2

Part C 4

Writes the solution in interval notation with square brackets to match the inclusive symbol. 2
Describes the number-line picture correctly, with the circles and shading matching the symbol and the chosen interval. 2

6. Two crossings is not a law 10 points

Part A 3

Chooses a specific quadratic and computes its discriminant correctly. 2
States that the count found contradicts the claim of exactly two. 1

Part B 3

Computes the discriminant and reads the intercept count from it. 2
Describes the vertex's position relative to the $x$ -axis, consistent with the count it found. 1

Part C 4

States the three-way discriminant classification completely. 2
Explains, with a case where $a > 0$ but the count is not two, why restricting to upward parabolas fails. 2

7. A pen with one side already built 10 points

Part A 3

Gets the parallel side by subtracting the two $x$ -sides from the total fence, not from a full four-sided perimeter. 2
Multiplies the two dimensions to write the area as a single quadratic in $x$ . 1

Part B 4

Finds the optimizing $x$ by applying $x = -\tfrac{b}{2a}$ to the area expression. 2
Substitutes that $x$ into the parallel-side expression, not the area formula, for the second dimension. 1
Reports the two side lengths, the dimensions asked for, not the area they enclose. 1

Part C 3

Computes the area from the two dimensions found in part B. 2
Explains that the area is the optimal value while the dimensions are the optimizing input, so the two are not interchangeable. 1

8. A price the parabola prefers 13 points

Part A 3

Builds the price by subtracting the drops and the quantity by adding $30$ per drop. 2
Multiplies the two to state the revenue as a single product. 1

Part B 6

Finds the optimizing $x$ by applying $x = -\tfrac{b}{2a}$ to the expanded revenue. 2
Converts $x$ into the actual price $20 - x$ , since price rather than $x$ is asked for. 1
Substitutes the optimizing $x$ back into $R$ for the maximum revenue. 2
Reports the revenue as a figure distinct from the price, not the two run together. 1

Part C 4

Computes the revenue at the manager's price correctly. 2
Explains that a downward parabola with a single peak makes any other price give a lower revenue. 2

9. The one constant that keeps a parabola above the axis 11 points

Part A 3

Writes the discriminant correctly in terms of $k$ . 2
States the no-real-roots condition as the discriminant being negative. 1

Part B 4

Solves the inequality for $k$ , handling the direction correctly. 2
Reports the solution in $k$ with the boundary strictness handled correctly. 2

Part C 4

Substitutes the boundary value into the original quadratic and simplifies it correctly. 2
Explains why the boundary value fails the strict requirement, justifying its exclusion. 2

10. How high, and whether it clears the mark 11 points

Part A 3

Finds the peak time from $-\tfrac{b}{2a}$ . 2
Substitutes back for the peak height and labels which value is the input and which the height. 1

Part B 4

Forms the equation $h = 80$ , simplifies it, and computes its discriminant. 2
Reads the verdict from the sign of the discriminant. 2

Part C 4

Connects the peak height to the discriminant sign found in part B. 2
Explains that the peak time depends only on $a$ and $b$ while the peak height additionally depends on $c$ . 2

Multiple choice

Free response