Describe that parabola!
How's it moving?
Where's the increase and decrease?
Standard form!
Use that vertical motion formula!
100

f(x)= 3x2

narrow

100

f(x) = x2+12

12 up

100

f(x)= -5x2+1

x      y     (x,y)

-2    10    (-2,10)

-1    15    (-1,15)

0      20   (0,20)

1      15   (1,15)

2      10    (2,10)

increasing when x<0

decreasing when x>0

100

Find the axis of symmetry:

x2-8x+17

x=4

100

Write a vertical motion model in the form:

h(t)= -16t2+v0t+h0 for the situation:

Initial velocity is 32 ft/s

Initial height is 20 ft

-16t2+32t+20

200

f(x)= 1/2x2

wide

200

f(x) = (x-1)2

1 right

200

f(x)= -7x2+1

x      y     (x,y)

5    10    (5,10)

4    15    (4,15)

3      20   (3,20)

2     15   (2,15)

1      10    (1,10)

increasing when x<3

decreasing when x<3

200

Find the axis of symmetry:

-x2-2x-2

x=-1

200

Write a vertical motion model in the form:

h(t)= -16t2+v0t+h0 for the situation:

Initial velocity is 120 ft/s

Initial height is 50 ft

-16t2+120t+50

300

f(x)= -12x2

down and narrow

300

f(x) = (x+10)2-3

10 left and 3 down

300

f(x) = -10x2+12

x      y     (x,y)

-2    10    (-2,10)

-1    11    (-1,11)

0      12   (0,12)

1      11   (1,11)

2      10    (2,10)

increasing when x<0

decreasing when x>0

300

Find the axis of symmetry:

-x2+6x-8

x=3

300

Write a vertical motion model in the form:

h(t)= -16t2+v0t+h0 for the situation:

Initial velocity is 32 ft/s

Initial height is 20 ft

How many seconds will it take the object thrown to reach maximum height (Hint: are you finding the x or y value?)

-16t2+32t+20

x=1

It will take 1 second to reach the maximum height

400

f(x)= 0.75x2

wide

400

f(x) = (x+9)2+13

9 left and 13 up

400

f(x) = 3x2+8

x      y     (x,y)

-2    -10    (-2,-10)

-1    -15    (-1,-15)

0      -20   (0,-20)

1      -15   (1,-15)

2      -10    (2,-10)

increasing when x>0

decreasing when x<0

400

Find the VERTEX:

-3x2+6x

(1,3)

400

Write a vertical motion model in the form:

h(t)= -16t2+v0t+h0 for the situation:

Initial velocity is 32 ft/s

Initial height is 20 ft

What is the max height of the object thrown? (Hint: are you finding the x or y value?)

-16t2+32t+20

y= 36

The max height is 36 feet

500

f(x)= -0.3x2

down and wide

500

f(x) = (x-13)2-4

13 right and 4 down

500

f(x) = 9x2+6

x      y     (x,y)

15    -50    (15,-50)

10    -40     (10,-40)

5    -30       (5,-30)

0      -40     (0,-40)

-5      -50    (-5,-50)

increasing x>5

decreasing x<5

500

Find the VERTEX:

-2x2-16x-31


(-4,1)

500

Write a vertical motion model in the form:

h(t)= -16t2+v0t+h0 for the situation:

Initial velocity is 64ft/s

Initial height is 30 ft

How many seconds will it take the object thrown to reach maximum height? What is the max height?

-16t2+64t+30

x= 2 (2 seconds to reach max height)

y= 94 (94 ft is max height)



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