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shannon brown

on 14 January 2013

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Algebra 2 Quadratic Equations By: Danielle Nasuti, Nicholas Syrylo, Shannon Brown, Xena Karalis, and Stephen Pinone.
Math isn't just used in the classroom; it is used in every aspect of our daily lives, even quadratic functions. Quadratics focus on parabolas, and these parabolas are often used in sports such as archery, baseball, javelin, basketball, volleyball, and many more to find things like distance or time. Volleyball Suppose a volleyball player serves from 1 meter behind the back line. If no other player touches the ball, it will land in bounds. The equation h = -4.9t² + 3.82t + 1.7 gives the ball’s height (h) in meters in terms of time (t) in seconds.What time will the ball land? To solve this problem, we could use the quadratic formula to figure out the time. A= -4.9
B=3.82
C= 1.7 h = -4.9t² + 3.82t + 1.7 Equations Standard Form:
Y = -4.9t² + 3.82t + 1.7 Vertex Form: (.38, 2.44) t, h h, k Y= -4.9(x-0.38)^2 + 2.44 Real World Problem Suppose that Tonya is a volleyball player and she serves from 2 meters behind the back line on the volleyball court. If no other player touches the ball, it will land in bounds. The equation h= -3x^2 + 8x + 3 gives the ball’s height h in meters in terms of time x in seconds. What time will the ball land? a=-3
b= 8
c=3 Work: h= -3x^2 + 8x + 3 x=-1/3 x=3 Actual Graph of y = -3x^2+8x+3 Equations in: Standard Form: Y= -3x^2 + 8x + 3 Vertex Form: a=-4.9
b=3.82
c=1.7 a=-3
b=8
c=3 (1 1/3, 8 1/3) X, Y h, k The ball will land at 3 seconds.
The vertex can represent where the volleyball begins to descend after reaching its maximum point.

The axis of of symmetry can represent the pull of gravity on the volleyball.

The x-intercepts represent where the the ball initially starts and where the ball finally lands.

The y-intercepts can represent how high the volleyball is in relation to its distance. It can also represent where the ball is served and hit. Soccer Example At a soccer game, a soccer ball is kicked into the air at a speed of 20 m/s. Its height, h, in meters, after t seconds is given by the formula: h = -5t² + 20t + 0. What time will the ball land? In order to find time, we could use the quadratic formula. Actual graph: h = -5t² + 20t + 0 a= -5
b= 20
c= 0 x= 4 x= (0,4) The ball will land at 4 seconds. Equations Standard Form: Y = -5t² + 20t + 0 Vertex Form: Actual Graph of y = -5x2+20x+0 a= -5
b= 20
c= 0 ( 2, 20) x, y h, k Y= -5(x-2)² + 20 Real World Problem Work Equations In a real world situation,

The vertex would represent the soccer ball’s maximum point in the air.
The axis of symmetry represents the pull of gravity.
The x-intercepts represent where the ball is initially kicked and where it lands on the ground.
The y-intercepts represent how high the ball was kicked in relation to its time and distance. So the next time you watch a game, you can think about how math and quadratic functions play a role in our lives when we least expect it! In the Soccer World Cup, Jerry kicks a ball across the field, headed toward the soccer net across from him. The function h = -4x^2 + 24x+ 1 models the height (h) of the soccer ball in meters as it travels in time (x) seconds. If there is no wind, and he kicks the ball in the air, what time will the ball be at its maximum point? What time will the ball land? h = -4x^2 + 24x+ 1 A=-4
B=24
C=1 x= To solve this real world problem, we would have to find the time in terms of x first, and then find the vertex. You can't have negative time. Standard Form: Y = -4x^2 + 24x+ 1 Vertex Form: We will be using the Quadratic Formula. A= -4
B= 24
C= 1 ( 3, 37) X Y h, k Y= -4(x-3)^2 + 37 Actual graph: See, math is used
in our everyday lives! (-0.04, 0) (6, 0) (0,0) (4, 0) t= -0.32 (1.1, 0) (-0.32,0) (-1/3,0) (3,0) You can also solve this problem by factoring! -5x2+20x+0 = -5(x+0)(x-4) -5(x+0)=0
-5x+0=0
-5x=0
0/-5 = -0 -5(x-4)=0
-5x+20=0
-5x=-20
-20/-5 =4 The best part about quadratic functions is that you can use any of the formulas, such as completing the square, the quadratic formula, and factoring, to get the same answers! (3, 37) To find the balls maximum point, we would have to find the vertex. At 3 seconds, the ball will be at a maximum height of 36 meters.
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