Zeros: 1, 3x x

2( ) ( )P x a x h k cbxaxxP 2)(

2 2Write ( ) 2 4 16 in the form ( ) ( ) .P x x x P x a x h k Example 1

1642)( 2 xxxP

Divide by 2 to make the coefficient of x2 equal to 1.

822

)( 2 xxxP

xxxP 282

)( 2

21( ) 8 22

1P x x x

Add 8 to both sides.

Add [½·2]2 = 1 to both sides to “Complete the Square” on the right.

2)1(92

)( xxP Combine terms on the left; factor

on the right.

9)1(2

)( 2 xxPSubtract 9 from both sides.

18)1(2)( 2 xxP Multiply both sides by 2.

From we can determine several components of the graph of

,18)1(2)( 2 xxP.1642)( 2 xxxP

2, 1, 18

vertex: ( , ) ( 1, 18)

Axis of Symmetry : 1

Domain: ( , ), Range: [ 18, )

Decreasing on ( , 1)

Increasing on ( 1, )

a h k

h k

x

x y

x

x

One method to determine the coordinates of the vertex is to complete the square.

Rather than go through the procedure for each individual function, we generalize the result for P(x) = ax² + bx + c.

The graph of

(a) is a parabola with vertex (h,k), and the vertical line x = h as axis of symmetry;

(b) opens upward if a > 0 and downward if a < 0;

(c) is wider than and narrower than

,0,)()( 2 akhxaxP

10 if 2 axy.1 if 2 axy

Vertex Formula for Parabola P(x) = ax² + bx + c (a 0)

abac

abxaxP

aacb

abx

ay

abx

aacb

ay

abx

abx

ab

ac

ay

xabx

ac

ay

acx

abx

ay

acbxaxyacbxaxxP

44

2)(

44

2

244

44

)0()0()(

22

2

22

2

2

2

2

22

2

2

2

2

2

2

Standard form

Replace P(x) with y to simplify notation.

Divide by a.

Subtract

Add

Combine terms on the left; factor on the right.

.ac

.42

1

2

22

a

b

a

b

Get y-term alone on the left.

Multiply by a and write in the form .)()( 2

khxaxP

h

k

Height of a Propelled Object

The coefficient of t ², 16, is a constant based on gravitational force and thus varies on different surfaces.Note that s(t) is a parabola, and the variable x will be used for time t in graphing-calculator-assisted problems.

Height of a Propelled Object

If air resistance is neglected, the height s (in feet) of an object propelled directly upward from an initial height s0 feet with initial velocity v0 feet per second is

where t is the number of seconds after the object is propelled.

00216)( stvtts

A ball is thrown directly upward from an initial height of 100 feet with an initial velocity of 80 feet per second.(a) Give the function that describes height in terms of time t.

(b) Graph this function.

(c) The cursor in part (b) is at the point (4.8,115.36). What does this mean?

1008016)( 2 ttts

After 4.8 seconds, the object will be at a height of 115.36 feet.

Example 2

(d) After how many seconds does the projectile reach its maximum height?

(e) For what interval of time is the height of the ball greater than 160 feet?

ft. 200 ofheight a reaches projectile theseconds, 2.5After

.200100)5.2(80)5.2(16 and

5.2)16(2

80

2

vertexat the occurs maximum The

2

y

a

bx

Using the graphs, t must be between .919 and 4.081 seconds.

(f) After how many seconds will the ball hit the ground?

When the ball hits the ground, its height will be 0, so we need to find the positive x-intercept. From the graph, the x-intercept is about 6.036, so the ball will reach the ground 6.036 seconds after it is projected.