Interpreting the Vertex of a Projectile Model
Mr. Delgado's Algebra I class of 26 is analyzing the height of a ball thrown upward. The function is h(t) = −16t² + 64t + 5, where h is feet and t is seconds. During board work, one student says the maximum height occurs at t = 5 seconds because “5 is the constant and it is the highest number in the expression.” Another says the maximum is at t = 64 seconds because “64t is the biggest term.” A third correctly computes t = 2 but cannot explain why the vertex formula works. Mr. Delgado has one class period before a quiz that includes a similar projectile item. He wants students to connect the algebraic form to the graph and to the physical situation without simply memorizing t = −b/(2a).
1. Diagnose the two incorrect responses and the third student's incomplete explanation. What does each reveal about understanding of quadratic structure?
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- t = 5: The student treats the constant term as the location of the maximum. This confuses the y-intercept with the vertex. The constant 5 is the initial height at t = 0, not the time of maximum height.
- t = 64: The student treats the linear coefficient as the time of the maximum. The coefficient 64 is the initial upward velocity in feet per second; it affects the vertex but is not the vertex's t-coordinate.
- t = 2 with no explanation: The student can apply t = −b/(2a) procedurally but may not see that the vertex lies on the axis of symmetry x = −b/(2a) or that the two roots are equidistant from it. This is fragile knowledge that may not transfer to factored or vertex forms.
The pattern shows that students are attending to individual numbers rather than the structure of the quadratic. The lesson should connect the coefficients to graph features and to the physical context.
2. Write a concise explanation for the class that connects the vertex, symmetry, and the physical situation. Include a table or graph description.
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For h(t) = −16t² + 64t + 5, the graph is a downward-opening parabola because the coefficient of t² is negative. The maximum height occurs at the vertex. Using t = −b/(2a): t = −64/(2 × −16) = 2 seconds. At t = 2, h(2) = −16(4) + 64(2) + 5 = 69 feet.
A table shows the symmetry: t = 0, h = 5; t = 1, h = 53; t = 2, h = 69; t = 3, h = 53; t = 4, h = 5. The heights at t = 1 and t = 3 are equal, and the heights at t = 0 and t = 4 are equal, so the vertex is halfway between symmetric time values. Physically, the ball starts at 5 feet, rises for 2 seconds, reaches 69 feet, then falls symmetrically. The constant 5 is the starting height, and 64 is the initial upward velocity, not the time of the maximum.
3. Propose one quiz item that distinguishes memorized formula use from conceptual understanding. Include expected responses.
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Quiz item: “A ball's height is h(t) = −16t² + 48t + 4. Without using the formula t = −b/(2a), determine the time of maximum height and explain using symmetry.”
Conceptual response: The student builds a table: t = 0, h = 4; t = 1, h = 36; t = 2, h = 36; t = 3, h = 4. Since h(1) = h(2), the vertex is halfway between 1 and 2, at t = 1.5 seconds. The student explains that a parabola is symmetric about its vertex. Procedural-only response: The student immediately writes t = −48/(2 × −16) = 1.5 but cannot explain why or connect it to equal heights. Misconception response: The student says the maximum is at t = 4 because 4 is the constant. The item separates students who can reason from the graph/table from those who only memorized a formula.