Parabolic Motion Math 112 Instructional Objectives…….…..…..3 Driving Question…………………………..4 Maximum Height and Time……….….5 Completing the...

QUADRATICS

Parabolic MotionMath 112

TABLE OF CONTENTSInstructional Objectives…….…..…..3Driving Question…………………………..4Maximum Height and Time……….….5Completing the Square………………..6Coordinates of Vertex…………….…….9Graphing Technology:Table……..…10

Graph……………………………....11Methods to Solve for Height……….12

Transformational Form…………….13General Form………………………….…15Quadratic Formula………….…16Graphing Technology……………..…17

INSTRUCTIONAL OBJECTIVES

C9 – translate between different forms of quadratic functions

C23 – solve problems involving quadratic equations in a variety of ways

C31 – analyze and describe the characteristics of quadratic functions

DRIVING QUESTIONSA baseball is batted into the air and follows the parabolic path defined by the equation

h = - 2t2+16t+1,

where t is time in seconds and h is height in feet.

A) What is the maximum height reached by the ball, and at what time does it reach maximum height?

B) When is the ball at a height of 25 feet?

MAXIMUM HEIGHT AND TIME

We will answer question A by finding the

vertex. We will focus on two methods to

determine the coordinates of the vertex:

1) Write the equation in transformational form by completing the square.

2) Use graphing technology.

COMPLETING THE SQUARE

Begin with the equation in general form as it was given.

h = - 2t2 + 16t + 1

Step 1: Move all constants to the left side.h – 1 = - 2t2 + 16t

Step 2 : Factor out the coefficient of t2 from both terms involving t variable

h – 1 = - 2(t2 - 8t)

COMPLETING THE SQUARE(CON’T)

Step 3: Determine what value must be added in to create a perfect square on the right side.

- 2(t2 - 8t +16)

Step 4: Balance the equation by adding an equivalent amount to the left side.

h – 1- 32 = - 2(t2 - 8t + 16)

Divide the factored “b” term by 2 and square it

Step 5: Factor the perfect square.h - 33 = -2(t - 4)2

Step 6: Rearrange the formula into transformational form.

-1/2(h - 33) = (t - 4)2

COMPLETING THE SQUARE(CON’T)

COORDINATES OF THE VERTEX

-1/2(h - 33) = (t - 4)2

The vertex which occurs at a maximum here has coordinates (4, 33)

4 is in brackets with t, therefore represents time when the ball is at maximum height.

33 is in brackets with h, therefore represents the maximum height.

Therefore the ball reachesa maximum height of 33 feetat 4 seconds.

GRAPHING TECHNOLOGY (OPTION 1)

Because we know the equation is quadratic and we see symmetry in the table, then we can be sure that the vertex has coordinates (4, 33). 4 is an x-value which represents time, and 33 is a y-value which represents height.

Therefore the ball reaches a maximum height of 33 feet at 4 seconds.

1. Insert the equation into [Y=]

2. Generate a table of values to determine the greatest y-value, and its corresponding x-value.

X Y

1 15

2 25

3 31

4 33

5 31

GRAPHING TECHNOLOGY (OPTION 2)

Therefore the ball reaches a maximum height of 33 feet at 4 seconds.

1. Insert the equation y = - 2x2 + 16x + 1into [Y=]

2. Adjust the windowsettings so the graphcan be viewed on thescreen. Then press GRAPH.Let’s do it together!

3. Use the CALCULATE feature to set the left andright boundaries for the maximum. The calculator willgenerate the coordinates for the maximum. Let’s try.

HEIGHT OF 25 FEET?

Transformational Form

General Form

Graphing Calculator

TRANSFORMATIONAL FORM

-1/2(h - 33) = (t - 4)2

Step 1: Substitute h as 25 and simplify the left side of the equation.

-1/2(25 - 33) = (t - 4)2

-1/2(-8) = (t - 4)2

4 = (t - 4)2

Step 2: Square root both sides.+/- 2 = t – 4

Step 3: Add 4 to both sides.+/- 2 + 4 = t

CONSIDER BOTH CASES

+2 + 4 = t6 = t

The ball is at a height of 25 feet at 6 seconds.

-2 + 4 = t2 = t

The ball is at a height of 25 feet at 2 seconds

Does it make sense that there are two answers to this question? Explain.

GENERAL FORMh = - 2t2+16t+1

Because we want to know the time the ball is at a height of 25 feet, we must:

substitute 25 as h, set the left side equal to zero, and solve using the quadratic formula.

25 = - 2t2+16t+1

0 = - 2t2 + 16t - 24For a closer look how to apply

the quadratic formula, let’s view this short explanation.

QUADRATIC FORMULA

Substitute the values of a, b, and c into the formula above.

Again there are two solutions:

x = (-16+8)/(-4) OR x = (-16-8)/(-4)x = (-8)/(-4) x = (-24)/(-4)x = 2 x = 6

Therefore the ball reaches a height of 25 feet at2 seconds and at 6 seconds.

a = -2b = 16c = -24

GRAPHING TECHNOLOGY

We can use the table feature in much the same way we did to determine the coordinates of the vertex.

1. Write the equation in [Y =]

2. Generate the table.

3. Locate all y-values of 25 in

the table.

4. Read the corresponding x-

values when y=25.In the table we see the ordered pair (2, 25) and (6, 25), therefore the ball reaches a height of 25 feet at 2 seconds and 6 seconds.

X Y

2 25

3 31

4 33

5 31

6 25