Recognizing Quadratic Functions and Interpreting Graphs
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AIM

How does recognizing the shape and key features of a parabola help us describe real-world motion and make predictions?

Objectives

Tier 1: High Support

1. Look at the graph of$y = x^2 - 4$shown below. Identify the vertex and the zeros (where the graph crosses the x-axis). [GRAPH: Coordinate plane showing parabola$y = x^2 - 4$for$-4 \leq x \leq 4$and$-5 \leq y \leq 13$. Grid lines present. Vertex labeled at$(0, -4)$. Zeros marked at$(-2, 0)$and$(2, 0)$. Axes labeled with x-axis from -4 to 4 and y-axis from -5 to 13.]
The vertex is the lowest point on this parabola, located at the ordered pair ______ , ______.
The zeros are where the graph crosses the x-axis, which happens at ______ and ______.
The vertex tells us that the minimum y-value is ______ when$x =$______.
This parabola opens upward because the coefficient of$x^2$is ______ (positive or negative).
2. For the function$y = x^2 - 4$, identify the domain and range. Use the graph to justify your answer. [GRAPH: Same coordinate plane as Problem 1, showing parabola$y = x^2 - 4$with vertex at$(0, -4)$and zeros at$(-2, 0)$and$(2, 0)$.]
The domain is all possible ______ -values. For this parabola, x can be any real number, so domain = ______ .
The range is all possible ______ -values. The parabola's lowest point is at$y = -4$, so$y \geq$______ .
Looking at the graph, the parabola opens ______ and never goes below the vertex.
In interval notation, the range is ______ ∞ ) or in inequality form,$y \geq$______ .
3. A basketball player shoots a ball that follows a parabolic path. The height$h$(in meters) of the ball at time$t$(in seconds) is given by$h(t) = -5t^2 + 10t + 2$. Which of the following is the best description of the graph of this function? (A) Opens upward with a maximum height of 7 meters (B) Opens downward with a maximum height of 7 meters (C) Opens upward with a minimum height of 2 meters (D) Opens downward with a minimum height of 2 meters [VISUAL: Illustration of a basketball player shooting a ball in a parabolic arc, with the ball at various points along the curved path.]
The coefficient of$t^2$is ______ , which is negative, so the parabola opens ______ .
When a parabola opens downward, it has a ______ (maximum or minimum) point at the vertex.
For a real-world ball toss, the graph starts at ground level or above, rises to a peak, then ______ .
The correct answer is (______) because the parabola opens ______ and the highest point is called a maximum.
4. Look at the four graphs below. Three graphs represent quadratic functions. One graph is NOT a quadratic function. Identify which graph is not quadratic and explain why. [GRAPH SET: Four coordinate planes labeled A, B, C, D: A: Parabola opening upward with vertex at (1, 0) B: Straight line with slope 2 passing through origin C: Parabola opening downward with vertex at (-1, 3) D: Parabola opening upward with vertex at (0, -2)] All axes labeled$-5$to$5$.
Graphs A, C, and D show ______ shapes (curved/straight). Graph B shows a ______ shape (curved/straight).
A quadratic function has a graph that is always a ______ , which is U-shaped or upside-down U-shaped.
Graph B is not a quadratic function because it is a ______ line, not a parabola.
The answer is ______ because it does not have the curved parabolic shape required for a quadratic function.

Tier 2: Standard

1. The graph of$y = x^2 - 4$is shown below. Identify the vertex, the zeros (x-intercepts), and explain why the y-intercept is located at$(0, -4)$. [GRAPH: A parabola opening upward. The vertex is at$(0, -4)$. The graph extends from$x = -4$to$x = 4$on the x-axis and from$y = -5$to$y = 13$on the y-axis. Grid lines are visible. The parabola intersects the x-axis at$(-2, 0)$and$(2, 0)$. The parabola intersects the y-axis at$(0, -4)$. The vertex is marked with a dot and labeled. The zeros are marked with dots and labeled.]
2. Use the graph of$y = x^2 - 4$to determine the domain and range of this function. Justify your answers by explaining what domain and range represent.
3. A ball is kicked from ground level and follows a parabolic path modeled by the equation$h(t) = -16t^2 + 32t$, where$h$is height in feet and$t$is time in seconds. Find the zeros of this function and interpret what each zero represents in the context of the ball's motion. [ILLUSTRATION: A sketch of a ball's parabolic trajectory. The ball starts at ground level (origin), rises to a maximum height, and returns to ground level. The path is labeled with the vertex marking the maximum height and the zeros marking where the ball leaves the ground and where it lands.]
4. Three graphs are shown below. Graph A shows$y = x^2 - 4$, Graph B shows$y = x^2 + 1$, and Graph C is incorrectly labeled as$y = x^2 - 4$but has a vertex at$(0, 4)$. Identify which graph is incorrect and explain why the vertex location contradicts the equation. [GRAPH A: A parabola opening upward with vertex at$(0, -4)$, zeros at$(-2, 0)$and$(2, 0)$. Grid and axes are labeled.] [GRAPH B: A parabola opening upward with vertex at$(0, 1)$, no x-intercepts visible. Grid and axes are labeled.] [GRAPH C: A parabola opening upward with vertex at$(0, 4)$, labeled as$y = x^2 - 4$. Grid and axes are labeled.]

Tier 3: Extension

1. A basketball player shoots a ball that follows a parabolic path modeled by the function$h(t) = -16t^2 + 32t + 6$, where$h$is the height in feet and$t$is time in seconds. (a) Analyze the vertex of this quadratic and explain what it represents in the context of the basketball shot. (b) Evaluate whether the ball will reach a height of 25 feet. Justify your reasoning using the quadratic formula or by completing the square. (c) Synthesize the features of the parabola (vertex, zeros, axis of symmetry, and direction of opening) to propose a real-world constraint that would make this model invalid for all positive values of$t$.
2. Consider the graph of$y = x^2 - 4$shown below. Your classmate claims that the range of this function is$y \ge -4$because 'the vertex is at the minimum, so all outputs are greater than or equal to the y-coordinate of the vertex.' (a) Evaluate your classmate's claim and justify whether it is correct or incorrect using properties of parabolas. (b) Address the counterargument that 'the range could be all real numbers because the parabola goes up forever.' Explain why this counterargument is valid or invalid. (c) Write a brief explanation (3-4 sentences) that you could give to your classmate to help them understand the correct range and why the vertex location determines it.
3. A school fundraiser sells T-shirts. The profit function is$P(x) = -2(x - 30)^2 + 1800$, where$x$is the number of shirts sold and$P(x)$is the profit in dollars. (a) Analyze the vertex and zeros of this quadratic to determine the break-even points and maximum profit. (b) Evaluate the validity of the statement: 'If we sell 60 shirts, we will make more profit than if we sell 50 shirts, because 60 is closer to the vertex.' Use calculations to justify your answer. (c) Propose a domain restriction for this function that reflects real-world constraints. Justify your restriction by considering both the mathematical features of the parabola and practical business considerations (e.g., inventory, market size).
4. Below is the graph of$y = x^2 - 4$for$-4 \le x \le 4$. Your task is to identify and explain why the following statement is false: 'The domain of this function is$-4 \le x \le 4$because those are the$x$-values shown on the graph.' Use the graph and your understanding of domain and range to justify your answer. Then, propose what the actual domain and range of$y = x^2 - 4$are, assuming no additional restrictions are given.

Answer Key

Tier 1: High Support

1.
Look at the graph of$y = x^2 - 4$shown below. Identify the vertex and the zeros (where the graph crosses the x-axis). [GRAPH: Coordinate plane showing parabola$y = x^2 - 4$for$-4 \leq x \leq 4$and$-5 \leq y \leq 13$. Grid lines present. Vertex labeled at$(0, -4)$. Zeros marked at$(-2, 0)$and$(2, 0)$. Axes labeled with x-axis from -4 to 4 and y-axis from -5 to 13.]
Answer: The vertex is at$(0, -4)$. This is the lowest point on the parabola. The zeros are at$x = -2$and$x = 2$. These are the points where the graph crosses the x-axis, meaning$y = 0$at those locations. We can verify: when$x = -2$,$y = (-2)^2 - 4 = 4 - 4 = 0$. When$x = 2$,$y = (2)^2 - 4 = 4 - 4 = 0$. The parabola opens upward because the coefficient of$x^2$is positive (it is$+1$).
High-support students often confuse the vertex with one of the zeros. Reinforce that the vertex is ONE point with both an x-coordinate and y-coordinate. Zeros are x-intercepts only (y = 0 always). Common error: reading$(0, -4)$as two separate zeros instead of one turning point.
2.
For the function$y = x^2 - 4$, identify the domain and range. Use the graph to justify your answer. [GRAPH: Same coordinate plane as Problem 1, showing parabola$y = x^2 - 4$with vertex at$(0, -4)$and zeros at$(-2, 0)$and$(2, 0)$.]
Answer: Domain: All real numbers, or$\mathbb{R}$, or$(-\infty, \infty)$. The graph extends left and right infinitely. Range:$y \geq -4$, or in interval notation$[-4, \infty)$. The parabola opens upward and has a minimum at the vertex, which is at$y = -4$. No y-values below -4 appear on the graph, but the parabola reaches all y-values from -4 and above.
Students often write the range incorrectly by not recognizing the minimum point. Common error: writing range as$(-4, \infty)$with a parenthesis instead of a bracket at -4. Since$y = -4$IS achieved at the vertex, we use a bracket:$[-4, \infty)$.
3.
A basketball player shoots a ball that follows a parabolic path. The height$h$(in meters) of the ball at time$t$(in seconds) is given by$h(t) = -5t^2 + 10t + 2$. Which of the following is the best description of the graph of this function? (A) Opens upward with a maximum height of 7 meters (B) Opens downward with a maximum height of 7 meters (C) Opens upward with a minimum height of 2 meters (D) Opens downward with a minimum height of 2 meters [VISUAL: Illustration of a basketball player shooting a ball in a parabolic arc, with the ball at various points along the curved path.]
Answer: The correct answer is (B) Opens downward with a maximum height of 7 meters. The coefficient of$t^2$is$-5$, which is negative. When the leading coefficient is negative, the parabola opens downward and has a maximum value at its vertex. We can find the vertex using$t = -\frac{b}{2a}$. Here,$a = -5$and$b = 10$, so$t = -\frac{10}{2(-5)} = -\frac{10}{-10} = 1$. At$t = 1$,$h(1) = -5(1)^2 + 10(1) + 2 = -5 + 10 + 2 = 7$. So the maximum height is 7 meters at time$t = 1$second. This makes physical sense: the ball is launched, reaches a peak of 7 meters, then falls back down.
Common error: Students choose (A) because they see 'maximum' and 'height' without checking the sign of the leading coefficient. Reinforce: NEGATIVE leading coefficient = opens DOWN = has a maximum. POSITIVE leading coefficient = opens UP = has a minimum. The real-world context (a ball path) should also guide intuition.
4.
Look at the four graphs below. Three graphs represent quadratic functions. One graph is NOT a quadratic function. Identify which graph is not quadratic and explain why. [GRAPH SET: Four coordinate planes labeled A, B, C, D: A: Parabola opening upward with vertex at (1, 0) B: Straight line with slope 2 passing through origin C: Parabola opening downward with vertex at (-1, 3) D: Parabola opening upward with vertex at (0, -2)] All axes labeled$-5$to$5$.
Answer: Graph B is NOT a quadratic function. Graphs A, C, and D all show parabolas (curved, U-shaped or inverted U-shaped), which are characteristic of quadratic functions. Graph B shows a straight line, which is a linear function (degree 1), not a quadratic function (degree 2). Quadratic functions always produce parabolic graphs; linear functions produce straight lines. Since Graph B is a line, it cannot represent a quadratic function.
Common error: Students may confuse 'parabola' with other curve shapes or not recognize that straight lines are never quadratic. Reinforce the visual: quadratic = always curved like a U or ⋂. Line = always straight = NOT quadratic. Use the word bank term 'parabola' consistently.

Tier 2: Standard

1.
The graph of$y = x^2 - 4$is shown below. Identify the vertex, the zeros (x-intercepts), and explain why the y-intercept is located at$(0, -4)$. [GRAPH: A parabola opening upward. The vertex is at$(0, -4)$. The graph extends from$x = -4$to$x = 4$on the x-axis and from$y = -5$to$y = 13$on the y-axis. Grid lines are visible. The parabola intersects the x-axis at$(-2, 0)$and$(2, 0)$. The parabola intersects the y-axis at$(0, -4)$. The vertex is marked with a dot and labeled. The zeros are marked with dots and labeled.]
Answer: Vertex:$(0, -4)$(the lowest point on the parabola, also called the minimum). Zeros (x-intercepts):$x = -2$and$x = 2$(where the parabola crosses the x-axis). The y-intercept occurs when$x = 0$. Substituting into the equation:$y = (0)^2 - 4 = -4$. Therefore, the y-intercept is$(0, -4)$, which is also the vertex in this case because the parabola is symmetric about the line$x = 0$.
Common error: Students may confuse the vertex with the y-intercept in all cases. Clarify that the y-intercept is always found by setting$x = 0$, while the vertex is the turning point and is found at the axis of symmetry. In this example they coincide, but this is not always true.
2.
Use the graph of$y = x^2 - 4$to determine the domain and range of this function. Justify your answers by explaining what domain and range represent.
Answer: Domain: All real numbers, or$(-\infty, \infty)$. Justification: The parabola extends infinitely to the left and right. For any real number$x$that you input, the function produces a valid output. Range:$[-4, \infty)$or$y \geq -4$. Justification: The parabola opens upward with a minimum point (vertex) at$(0, -4)$. The y-values start at$-4$and increase without bound as$x$moves away from the vertex in either direction. The parabola never goes below$y = -4$.
Common error: Students may state the range as$(-\infty, \infty)$if they fail to recognize that the parabola has a minimum value. Emphasize that the minimum y-value on the graph is$-4$(the vertex), so the range cannot include values below this.
3.
A ball is kicked from ground level and follows a parabolic path modeled by the equation$h(t) = -16t^2 + 32t$, where$h$is height in feet and$t$is time in seconds. Find the zeros of this function and interpret what each zero represents in the context of the ball's motion. [ILLUSTRATION: A sketch of a ball's parabolic trajectory. The ball starts at ground level (origin), rises to a maximum height, and returns to ground level. The path is labeled with the vertex marking the maximum height and the zeros marking where the ball leaves the ground and where it lands.]
Answer: To find the zeros, set$h(t) = 0$:$-16t^2 + 32t = 0$. Factor out$-16t$:$-16t(t - 2) = 0$. This gives$t = 0$or$t = 2$. The zeros are$t = 0$seconds and$t = 2$seconds. Interpretation: At$t = 0$, the ball is at ground level (initial position). At$t = 2$, the ball returns to ground level after reaching a maximum height. The ball is in the air for 2 seconds.
Common error: Students may forget to set the function equal to zero or may fail to factor correctly. Remind them that zeros are the values where the output equals zero (in this case, when height equals zero, or ground level). Also note that only the positive time values make sense in this context; negative time is not physically meaningful.
4.
Three graphs are shown below. Graph A shows$y = x^2 - 4$, Graph B shows$y = x^2 + 1$, and Graph C is incorrectly labeled as$y = x^2 - 4$but has a vertex at$(0, 4)$. Identify which graph is incorrect and explain why the vertex location contradicts the equation. [GRAPH A: A parabola opening upward with vertex at$(0, -4)$, zeros at$(-2, 0)$and$(2, 0)$. Grid and axes are labeled.] [GRAPH B: A parabola opening upward with vertex at$(0, 1)$, no x-intercepts visible. Grid and axes are labeled.] [GRAPH C: A parabola opening upward with vertex at$(0, 4)$, labeled as$y = x^2 - 4$. Grid and axes are labeled.]
Answer: Graph C is incorrect. If the equation is$y = x^2 - 4$, then the vertex should be at$(0, -4)$, not$(0, 4)$. To verify: when$x = 0$,$y = (0)^2 - 4 = -4$. Graph C shows a vertex at$(0, 4)$, which would correspond to an equation like$y = x^2 + 4$, not$y = x^2 - 4$. The constant term in the equation determines the vertical shift of the parabola. A negative constant ($-4$) shifts the parabola down 4 units, while a positive constant ($+4$) shifts it up 4 units.
Common error: Students may confuse the sign of the constant term and its effect on the vertex. Reinforce that the y-coordinate of the vertex for$y = a(x - h)^2 + k$is$k$. In standard form$y = x^2 - 4$, we can rewrite as$y = (x - 0)^2 - 4$, so the vertex is at$(0, -4)$, not$(0, 4)$.

Tier 3: Extension

1.
A basketball player shoots a ball that follows a parabolic path modeled by the function$h(t) = -16t^2 + 32t + 6$, where$h$is the height in feet and$t$is time in seconds. (a) Analyze the vertex of this quadratic and explain what it represents in the context of the basketball shot. (b) Evaluate whether the ball will reach a height of 25 feet. Justify your reasoning using the quadratic formula or by completing the square. (c) Synthesize the features of the parabola (vertex, zeros, axis of symmetry, and direction of opening) to propose a real-world constraint that would make this model invalid for all positive values of$t$.
Answer: (a) The vertex form can be found by completing the square or using$t = -\frac{b}{2a} = -\frac{32}{2(-16)} = 1$. At$t = 1$,$h(1) = -16(1)^2 + 32(1) + 6 = -16 + 32 + 6 = 22$feet. The vertex is$(1, 22)$, which represents the maximum height of 22 feet reached at 1 second after the shot. (b) Set$-16t^2 + 32t + 6 = 25$. Rearranging:$-16t^2 + 32t - 19 = 0$, or$16t^2 - 32t + 19 = 0$. Using the discriminant:$\Delta = (-32)^2 - 4(16)(19) = 1024 - 1216 = -192 < 0$. Since the discriminant is negative, there are no real solutions. The ball will not reach 25 feet because the maximum height is 22 feet, which is less than 25 feet. (c) The zeros of the function represent times when the ball is at ground level ($h = 0$). Solving$-16t^2 + 32t + 6 = 0$:$t = \frac{-32 \pm \sqrt{1024 - (-384)}}{-32} = \frac{-32 \pm \sqrt{1408}}{-32} \approx \frac{-32 \pm 37.53}{-32}$. This gives$t \approx 2.11$seconds (taking the positive root). A real-world constraint is that the parabola becomes invalid after the ball hits the ground or another surface; the model assumes the ball travels through empty space, but in reality, it cannot descend below ground level ($h = 0$). Therefore, the domain must be restricted to$0 \le t \le 2.11$seconds.
An exemplary response recognizes that the vertex represents the peak of the trajectory and correctly interprets the discriminant to determine feasibility. The student should justify why the ball cannot exceed 22 feet by comparing the given height to the vertex value. Part (c) demonstrates synthesis by connecting the mathematical features (zeros, parabolic shape) to a physical constraint, showing that models have limited domains in real-world contexts.
2.
Consider the graph of$y = x^2 - 4$shown below. Your classmate claims that the range of this function is$y \ge -4$because 'the vertex is at the minimum, so all outputs are greater than or equal to the y-coordinate of the vertex.' (a) Evaluate your classmate's claim and justify whether it is correct or incorrect using properties of parabolas. (b) Address the counterargument that 'the range could be all real numbers because the parabola goes up forever.' Explain why this counterargument is valid or invalid. (c) Write a brief explanation (3-4 sentences) that you could give to your classmate to help them understand the correct range and why the vertex location determines it.
Answer: (a) The classmate's claim is correct. The function$y = x^2 - 4$is a parabola that opens upward (since the coefficient of$x^2$is positive). The vertex is at$(0, -4)$, which is the minimum point on the graph. Since the parabola opens upward, all points on the curve have$y$-values greater than or equal to the minimum$y$-value at the vertex. Therefore, the range is$y \ge -4$or$[-4, \infty)$. (b) The counterargument is partially valid but incomplete. It is true that the parabola extends upward infinitely (to the right and left of the vertex), so outputs do increase without bound. However, this does not mean the range is all real numbers. The critical feature is that the parabola has a minimum value: no output is less than$-4$. The counterargument correctly identifies that$y$can be arbitrarily large, but it overlooks the lower bound created by the vertex. A more complete statement would be: 'The parabola goes up forever, so there is no upper limit to the range, but the vertex creates a lower limit.' (c) 'The range of a parabola is determined by the vertex, which is the lowest or highest point on the curve. In this function, the vertex at$(0, -4)$is the minimum point because the parabola opens upward. This means no output value can be less than$-4$, but the parabola continues upward forever, so larger values are always possible. Therefore, the range includes$-4$and all values above it:$y \ge -4$.'
An exemplary response demonstrates understanding of how the direction of opening and vertex location jointly determine the range. Students should recognize that the counterargument contains a kernel of truth (unbounded growth) but misses the essential constraint (the minimum). The written explanation in part (c) should be clear enough for a peer to understand, avoiding jargon and using the graph as a reference.
3.
A school fundraiser sells T-shirts. The profit function is$P(x) = -2(x - 30)^2 + 1800$, where$x$is the number of shirts sold and$P(x)$is the profit in dollars. (a) Analyze the vertex and zeros of this quadratic to determine the break-even points and maximum profit. (b) Evaluate the validity of the statement: 'If we sell 60 shirts, we will make more profit than if we sell 50 shirts, because 60 is closer to the vertex.' Use calculations to justify your answer. (c) Propose a domain restriction for this function that reflects real-world constraints. Justify your restriction by considering both the mathematical features of the parabola and practical business considerations (e.g., inventory, market size).
Answer: (a) The function is in vertex form$P(x) = -2(x - 30)^2 + 1800$. The vertex is$(30, 1800)$, meaning the maximum profit of$1800 dollars is achieved when 30 shirts are sold. To find break-even points, set$P(x) = 0$:$-2(x - 30)^2 + 1800 = 0 \Rightarrow (x - 30)^2 = 900 \Rightarrow x - 30 = \pm 30 \Rightarrow x = 0$or$x = 60$. The break-even points are at$x = 0$(no shirts sold, no profit or loss) and$x = 60$(selling 60 shirts results in zero profit). Between these points, profit is positive. (b) The statement is incorrect. While 60 is farther from the vertex than 50, we must calculate actual profits.$P(50) = -2(50 - 30)^2 + 1800 = -2(400) + 1800 = -800 + 1800 = 1000$dollars.$P(60) = -2(60 - 30)^2 + 1800 = -2(900) + 1800 = -1800 + 1800 = 0$dollars. Since$P(50) = 1000 > P(60) = 0$, selling 50 shirts yields more profit than selling 60 shirts. The statement confuses 'proximity to the vertex' with 'higher profit value'; being closer to the vertex does generally yield higher profit, but 60 is a break-even point where profit drops to zero. (c) A realistic domain restriction is$0 \le x \le 60$(and possibly tighter, e.g.,$0 \le x \le 50$). Mathematically, this range includes all break-even and profitable sales levels; profit is zero or positive only when$0 \le x \le 60$. Practically, a school would only want to sell shirts while generating profit or at least breaking even. Additionally, the school likely has inventory constraints (limited blank shirts available) and market constraints (limited demand from students and families), so setting an upper limit below 60 (such as 50 shirts) might be appropriate. For example,$0 \le x \le 50$ensures profit stays above$1000 dollars while accounting for potential supply chain challenges.
An exemplary response demonstrates facility with vertex form, accurate computation of break-even points, and recognition that proximity to the vertex does not automatically ensure higher profits (the shape of the parabola matters). Part (b) requires explicit calculation to refute the student's faulty reasoning. Part (c) shows synthesis by combining the mathematical constraints (zeros, profit positivity) with realistic business reasoning, demonstrating that domain restrictions in modeling problems reflect both mathematics and context.
4.
Below is the graph of$y = x^2 - 4$for$-4 \le x \le 4$. Your task is to identify and explain why the following statement is false: 'The domain of this function is$-4 \le x \le 4$because those are the$x$-values shown on the graph.' Use the graph and your understanding of domain and range to justify your answer. Then, propose what the actual domain and range of$y = x^2 - 4$are, assuming no additional restrictions are given.
Answer: The statement is false because the domain is not restricted to the$x$-values displayed on the graph. The domain of a function is the set of all possible input values for which the function is defined. The quadratic function$y = x^2 - 4$is a polynomial and is defined for all real numbers unless explicitly restricted. The graph shown only displays a window from$-4$to$4$on the$x$-axis for visualization purposes, but this does not mean the function only exists in that interval. For example,$y = (5)^2 - 4 = 21$is a valid output; the point$(5, 21)$is on the function even though it is not shown on the displayed graph. The actual domain of$y = x^2 - 4$is all real numbers, written as$(-\infty, \infty)$or$\mathbb{R}$, since any real number can be substituted for$x$and a valid$y$-value will result. The range, determined by the vertex$(0, -4)$and the upward-opening parabola, is$[-4, \infty)$or$y \ge -4$. The graph window (viewing window) is distinct from the domain and range; the viewing window is a choice made for clarity and convenience, while domain and range are mathematical properties of the function itself.
An exemplary response clearly distinguishes between the viewing window (the portion of the graph displayed) and the domain of the function (all allowable inputs). Students should recognize that a graph is a visual tool with practical limitations, but the mathematical function extends beyond what is shown. The response should use concrete examples (e.g., testing a point outside the window) to illustrate why the domain is larger than the displayed interval.