Vertical Motion Models
Parabolas
Calculating solutions to quadratics with the quadratic formula
Graphing radical functions
Simplifying radical expressions
100

If the vertex of a parabola is (265,89) what is the maximum height and the time at the max-height

Max height: 89

Time at the max-height: 

100

Calculate the vertex:

y=2x2=4x-3

(-1,-5)

100

y=2x2+5x+20

No Solution

100

y= √x

x    y

0    0

1    1

4    2

9     3

16    4

100

√45

3

200
What is the starting height h(t)=-6t2 +85t+42?


42

200

Identify the y-intercept

y=3x-6x+2

y-intercept:0,2

200

y=2x2-12x+18

1 Solution

200

y= √x+2

x     y

-2    0

-1    1

2     2

7     3

14   4

200

√140

2√35

300

How long would it take the object to hit the ground using h=-16t2=76t?

4.75

300

Identify the y-intercept and the point of symmetry:

y=2x2-8x-4

y-intercept: (0,-4)

point of symmetry: (4,-4)

300

2x2-5x=-7

No Solution

300

y= √x-3

x      y

3      0

4      1

7       2

12     3

19    4

300

√2 x √14

2√7

400

What are the Domain and range of the h=-16t2+76t?

Domain: 0 ≤ x ≤ 4.75

Range: 0 ≤ y ≤ 90.25

400

Using the vertex and the y-intercept, find the domain and range

Vertex: (2,-3)

Y-intercept: (0,1)

Domain: all real numbers

Range: y is greater than or equal to -3

400

5x2+21x=-18

x=-2,-1.2

400

y= √x-4

x      y

4     0

5      1

8      2

13    3

20   4

400

5√2 x√24

20√3

500

How long does it take the object to reach the max height? Using h=-16t2+76t

2.375seconds

500

Calculate the vertex and y-intercept

y=2-8x+6

Vertex=(2,-2)

y-intercept= (0,6)

500

x2-15=2x

5,-3

500

y=3 √x-2

x     y

4     0

5     1

8     2

13   3

20    4

500

√432

12√3

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