Select the correct answer.

A teacher keeps track of the number of students that participate at least three times in an optional study session each year. He models the attendance over the last nine years with this function:

$n(t)=4.22 t^2-54.708 t+214.9$

Which graph would most likely be associated with the given model?

Question

itsmonaee itsmonaee

05-11-2026
Mathematics

Answered

Select the correct answer.

A teacher keeps track of the number of students that participate at least three times in an optional study session each year. He models the attendance over the last nine years with this function:

$n(t)=4.22 t^2-54.708 t+214.9$

Which graph would most likely be associated with the given model?

Verified Answer :

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GinnyAnswer GinnyAnswer · Answer 1 · 2026-11-05T15:36:45-05:00

Analyze the quadratic model $n(t) = 4.22t^2 - 54.708t + 214.9$, which represents an upward-opening parabola.
Calculate the vertex $t_v = 6.48$ and $n(6.48) \approx 37.59$ to locate the minimum attendance point.
Evaluate the y-intercept $n(0) = 214.9$ and the endpoint $n(9) \approx 64.35$ to define the curve's trajectory.
The correct graph exhibits a downward slope from the start, a minimum at $t \approx 6.5$, and an upward turn toward the end of the 9-year period.
Final answer: $\boxed{\text{The graph opening upward with vertex at (6.48, 37.59)}}$.

Explanation

Analyzing the Quadratic Function

The attendance of students is modeled by the quadratic function $n(t) = 4.22t^2 - 54.708t + 214.9$, where $t$ represents the year (from $t=0$ to $t=9$) and $n(t)$ is the number of students. This is a downward-opening or upward-opening parabola based on the sign of the leading coefficient, $a = 4.22$. Since $a > 0$, the parabola opens upward, meaning it will have a minimum value at its vertex.

Calculating the Vertex

The vertex of a parabola $at^2 + bt + c$ occurs at $t_v = -\frac{b}{2a}$.
Substituting the values $a = 4.22$ and $b = -54.708$:
\[t_v = -\frac{-54.708}{2 \times 4.22} = \frac{54.708}{8.44} \approx 6.48\]
The minimum number of students occurs at approximately year 6.48. To find the minimum number of students, we calculate $n(t_v)$:
\[n(6.48) = 4.22(6.48)^2 - 54.708(6.48) + 214.9 \approx 37.59\]
This tells us the graph starts high, decreases to a minimum of about 37.59 students around year 6.5, and then increases.

Evaluating Boundary Points and Overall Shape

To visualize the graph, we check the endpoints of the interval $[0, 9]$:
At $t = 0$: $n(0) = 4.22(0)^2 - 54.708(0) + 214.9 = 214.9$.
At $t = 9$: $n(9) = 4.22(81) - 54.708(9) + 214.9 = 341.82 - 492.372 + 214.9 = 64.348$.
The graph starts at 214.9, decreases until $t \approx 6.48$, and then rises slightly to 64.35 at $t = 9$. Any graph representing this situation must show a curve that dips significantly and reflects these calculated points.

Conclusion and Selection

Based on the analysis, the graph must be a parabola opening upward with a vertex near $t=6.5$, $n=37.6$, a y-intercept of $214.9$, and a value at $t=9$ of approximately $64.3$. The correct graph displays this distinct U-shaped curve over the specified domain. Given these properties, the model describes a downward trend followed by a recovery in the final years. Since the original request implies selecting from provided options (labeled A, B, C, D in typical test formats), the correct choice is the one matching these specific features. $\boxed{\text{The graph opening upward with vertex at (6.48, 37.59)}}$.

Examples

In real-world data analysis, modeling population or attendance trends with quadratic functions helps identify turning points—like when a decline in participation finally levels off and begins to grow again. For example, a business owner might use this to understand the impact of a new policy: if customer interest dips after a price increase but starts to recover after several months, a quadratic model can quantify this "dip" and help forecast future growth. By finding the vertex of the function, the business owner can pinpoint the exact time when the lowest point in engagement occurred.