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Analyzing Functions Using Different Representations

7.14.3 Analyzing Functions Using Different Representations

7.14.3 • Analyzing Functions Using Different Representations

Activity

Here is a graph that represents the height of a baseball, h h , in feet as a function of time, t t , in seconds after it was hit by Player A.

A parabola on a coordinate grid. The x-axis represents the time in seconds and the y-axis represents the height in feet. The x-axis scale is 1 and extends from 0 to 7. The y-axis scale is 20 and extends from 0 to 120.

The function g g defined by g ( t ) = ( − 16 t − 1 ) ( t − 4 ) g ( t ) = ( − 16 t − 1 ) ( t − 4 ) also represents the height in feet of a baseball t t seconds after it was hit by Player B. Without graphing function g g , answer the following questions and explain or show how you know.

1.

Which player's baseball stayed in flight longer?

Player A's baseball stayed in flight longer. The zeros of function g g are 4 and − 1 16 − 1 16 . The negative zero doesn't have any meaning in this situation. A zero in this situation means the time when the baseball has a height of 0 (or when it hits the ground). For Player B, this happened when t = 4 t = 4 or 4 seconds after it was hit. Player A's baseball hit the ground a little over 5 seconds after it was hit.

2.

Which player's baseball reached a greater maximum height?

Player A's baseball reached a greater maximum height. From the graph for Player A, it looks like the y y -coordinate of the vertex is around 105 feet. For Player B, the vertex of the graph has a t t -coordinate of about 2 (halfway between − 1 16 − 1 16 and 4). We can find g ( 2 ) g ( 2 ) to estimate the height at that point. g ( 2 ) = ( − 16 ( 2 ) − 1 ) ( 2 − 4 ) = ( − 33 ) ( − 2 ) = 66 g ( 2 ) = ( − 16 ( 2 ) − 1 ) ( 2 − 4 ) = ( − 33 ) ( − 2 ) = 66 , so the highest point of Player B's baseball was around 66 feet.

3.

Based on the graph, what is the approximate height, in feet, of Player A’s baseball when hit? How do you know?

Compare your answer:

We can look at the y y -intercept of the graph, and it looks like the y y -coordinate is about 5.

4.

What is the height, in feet, of Player B’s baseball when hit?

4 feet. Find the height when t = 0 t = 0 . g ( 0 ) = ( − 16 ( 0 ) − 1 ) ( 0 − 4 ) = ( − 1 ) ( − 4 ) = 4 g ( 0 ) = ( − 16 ( 0 ) − 1 ) ( 0 − 4 ) = ( − 1 ) ( − 4 ) = 4 , so it was hit at 4 feet.

Video: Comparing Quadratic Functions

Watch the following video to learn more about comparing quadratic functions in different forms.

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Self Check

The functions describing the time that baseballs stayed in the air, in seconds, that four baseball players hit are below.

  • f ( t ) = ( − 16 t − 1 ) ( t − 5 )
  • g ( t ) = ( − 16 t − 1 ) ( t − 10 )
  • h ( t ) = ( t − 7 ) ( − 16 t − 1 )
  • p ( t ) = ( − 8 t − 1 ) ( 2 t − 5 )

Which function represents the ball that stayed in the air the longest?

  1. p ( t ) = ( − 8 t − 1 ) ( 2 t − 5 )
  2. h ( t ) = ( t − 7 ) ( − 16 t − 1 )
  3. g ( t ) = ( − 16 t − 1 ) ( t − 10 )
  4. f ( t ) = ( − 16 t − 1 ) ( t − 5 )

Additional Resources

Comparing the Graphs and Equations of Quadratic Functions

Jai kicks a football. The graph that represents the height of the football, h h , as a function of time, in seconds, is below.

A parabola on a coordinate grid. The x-axis scale is 1 and extends from negative 4 to 10. The y-axis scale is 10 and extends from approximately negative 45 to 80.

Kellan kicks a football and the height in feet, h h , is represented by the function h ( t ) = ( − 16 t − 1 ) ( t − 6 ) h ( t ) = ( − 16 t − 1 ) ( t − 6 ) .

Which player’s football stayed in the air longer?

Jai’s football landed after 4 seconds since the zero is at x = 4 x = 4 . The other zero, where x x is negative, does not make sense in this problem.

Looking at the function of Kellan’s ball, find the zeros by setting each factor equal to zero.

  • − 16 t − 1 = 0 − 16 t − 1 = 0 becomes t = − 1 16 t = − 1 16 . A negative time does not make sense in this problem.
  • t − 6 = 0 t − 6 = 0 becomes t = 6 t = 6 .

Kellan’s ball was in the air for 6 seconds, so his ball stayed in the air 2 seconds longer than Jai’s.

Try it

Comparing the Graphs and Equations of Quadratic Functions

Using the same information above, which player’s football reached a higher height?

Here is how to determine whose ball went higher:

Looking at the graph of Jai’s football, his football went about 70 feet high.

Step 1 - For Kellan’s football, find the vertex.
The vertex is halfway between the zeros. Since − 1 16 − 1 16 is near 0 and the other x x -intercept is at 6, we can use 3 as an estimate for the x x -coordinate of the vertex.

Step 2 - Substitute this into the function:
h ( t ) = ( − 16 ( 3 ) − 1 ) ( ( 3 ) − 6 ) = ( − 49 ) ( − 3 ) = 147 h ( t ) = ( − 16 ( 3 ) − 1 ) ( ( 3 ) − 6 ) = ( − 49 ) ( − 3 ) = 147 .

Kellan’s ball went 147 feet into the air so his football went higher.