1. Interpreting Quadratic Functions from Graphs and Tables
Eureka Math Algebra 1 Module 3 Lesson 10

2. Objectives
Students interpret quadratic functions from graphs and tables: zeros (𝑥-intercepts), 𝑦-intercept, the minimum or maximum value (vertex), the graph’s axis of symmetry, positive and negative values for the function, increasing and decreasing intervals, and the graph’s end behavior.
Students determine an appropriate domain and range for a function’s graph and when given a quadratic function in a context, recognize restrictions on the domain.
Standards
F.IF.B.4
F.IF.B.6

4. The interval from 0-6 seconds would be represent the dolphin jumping out of the water at 0 seconds and landing back in the water at 6 seconds.
Since the graph only records the vertical distance the dolphin is compared to the water we can not determine the horizontal distance.

5. The values of f (t) = 0 is when the dolphin is at the surface of the water. This happens at 0, 6, 16 and 24 seconds.
The dolphin is under water from the interval of 6 to 16 seconds. It is under water for a total of 10 seconds.

6. The maximum height is about 23 feet
The maximum height is about 23 feet. Since the y-axis represents the vertical height, we are looking for the highest point on the graph. This occurs roughly at (20,23).
f (t) = -50 means that the dolphin dived down and is 50 feet below the surface of the water. Looking at the graph it looks to appear at (11,-50) which means at 11 seconds of the video that the dolphin is 50 feet below the surface of the water.

8. The data can be represented using a quadratic function
The data can be represented using a quadratic function. There seems to be symmetry between the points where the vertex is at (6, 405)
Andrew initially invested $22,500. We get $22,500 by multiplying 225 by We can find that out by using the vertex and the symmetry of the graph. Since 6 months is the vertex, if we went back 6 months that would be 0 months or when he started his portfolio. This would be the same value as 12 months, moving 6 months ahead from the vertex.

9. The maximum value of his stock is $40,500. This happened at 6 months.
Andrew should sell his stock since after 6 months his portfolio begins to decline in value.

10. Between the months of 10 and 12, Andrew’s stock portfolio decreased a total of $10,000. Between the months of [12,14] he loses $14,000 so the average rate of change for [12,14] is even more.
No there is not, quadratic expressions have an exponential growth (or decay) so there will not be another time interval where Andrew’s stock is decreasing $10,000 in a two-month interval.