PARABOLASPARABOLASDo Now: Questions 1-9 find the equation of each parabola and name the transformation from x2

Things to know about Things to know about parabolasparabolas

 A parabola is the set of all points equidistant from a fixed line, called the Directrix, and a fixed point not on the line, called the focus.

  Equation of a parabola  y = a(x - h)2 + k  The vertex is at point (h, k)  The axis of symmetry is the line x = h  The graph of y = ax2 opens upward if a > 0 and downward if a <

0.  The larger the value of | a | is, the narrower the graph of the

parabola is.

TIME FOR THE HARD STUFFTIME FOR THE HARD STUFF

An object is launched at 19.6 meters per second (m/s) from a

58.8-meter tall platform. The equation for the object's height s at time t seconds after launch is

s(t) = –4.9t2 + 19.6t + 58.8, where s is in meters. When does

the object strike the ground?

STUDENT WORKSTUDENT WORK

ANSWERANSWER

0 = –4.9t2 + 19.6t + 58.8 0 = t2 – 4t – 12

0 = (t – 6)(t + 2)

The object strikes the ground six seconds after launch.

#2 #2

An object in launched directly upward at 64 feet per second (ft/s) from a platform 80 feet

high. What will be the object's maximum height? When will

it attain this height?

STUDENT WORKSTUDENT WORK

ANSWERANSWER

Equation: s(t) = –16t2 + 64t + 80

h = –b/2a = –(64)/2(–16) = –64/–32 = 2 k = s(2) = –16(2)2 + 64(2) + 80 = –16(4)

+ 128 + 80 = 208 – 64 = 144

It takes two seconds to reach the maximum height of 144 feet.

Notes On “Projectile Notes On “Projectile Problems” Problems”

s(t) represents the projectile's instantaneous height at any time t

vo represents initial velocity t represents time in seconds after the projectile is released

EXAMPLEEXAMPLE

A ball is thrown directly upward from an initial height of 200 feet with an initial

velocity of 96 feet per second. After how many seconds will the ball reach its maximum height? And, what is the

maximum height?

Pair SharePair Share

After Mrs. Dron got hit in the head with a pitch, Mr. Chester came up to bat. The equation h=-.004d2 +.96d+3 represents the path the ball took after Mr. Chester hit the first pitch. How far off the ground was the ball when it was hit by the bat?

AnswerAnswer

h=-.004d2 +.96d+3  h=-.004(0) + .96(0) + 3 h=  0 + 0 + 3 h= 3 feet, which is how far from the ground the ball was when it was hit by the bat

HomeworkHomework

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