Interpreting Quadratic Functions from Graphs

and TablesEureka Math Algebra 1 Module 3 Lesson 10

Objectives

• Students interpret quadratic functions from graphs and tables: zeros ( -𝑥intercepts), -intercept, the minimum or maximum value (vertex), the 𝑦graph’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

• 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.

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.

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.

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 100. 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.

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.

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.