QUADRATIC MODELS THROWING A BALL Quadratic Function fx
QUADRATIC MODELS THROWING A BALL
Quadratic Function f(x) = 2 ax + bx +c x intercepts (__, 0) y intercept (0 , __)
f(x) 2 =AX + BX +C If A is + Min If A is - Max
2 =2 X Ex. 1 f(x) -3 X -2 -2 Find Y intercept ( 0, __) Substitute 0 for X. 2 f(0) =2(0) -3(0) - 2 y = -2
Ex. 1 f(x) 2 =2 X -3 X -2 How do you find x intercepts? (___, 0 ) Let Y = 0 and solve for X. See book for Quadratic Formula Solution.
2 =2 X Ex. 1 f(x) -3 X -2 To find x intercepts? Solve by graphing. Use Calc. #2 Zero of f(x). left bound …
Or factor and let each factor = 0. 2 2 X - 3 X - 2 = 0 (2 X + 1 ) (X - 2) = 0 2 X + 1 = 0 or X - 2 = 0 2 X = -1 X=2 X = -1/2
Ex. 1 f(x) 2 =2 X -3 X -2 part b Since a is +, it’s min -b x value in vertex 2 a (-b/2 a, y)
Find ordered pair for vertex -b x value in vertex 2 a - (-3) = 3 (3/4, -3. 125 ___ ) 2(2) 4 get this by subst.
So what are the ways to solve? 2 f(x) 0 =2 X -3 X -2 Quadratic Formula Factor and let each factor = 0 Graph and look at the x interc (also called the zeros of f
Ex. 2 A ball is thrown up. Height is a function of time Given : ho = 15 m 15 Vo = 20 m/sec g = gravity (9. 8 m/s 2 or 2 32 ft/s )
h 0 = 15 Vo =20 h = - 1/2 g 2 t g =9. 8 + vot + ho 2 h = - 1/2 (9. 8) t + 20 t + 15 2 h = - 4. 9 t + 20 t + 15
b. Find ball height when t = 3 2 t + h = - 4. 9 t 20 t + 15 2 h = -4. 9( 3) 3 + 20(3) + 15 h = 30. 9 meters
When is the ball 25 m off ground? Ball is up 25 M so let h = 25 2 h = - 4. 9 t + 20 t + 15 2 + 25 = - 4. 9 t 20 t + 15
2 + t 25 = - 4. 9 20 t + 15 Put in standard form Get 0 on left. 2 t 0 = - 4. 9 + 20 t - 10 Solve with quadratic formula.
2 t 0 = - 4. 9 + 20 t - 10 2 0= Ax + Bx+ C -b + √ - 4 a c 2 a 2 b t = -20 + √ 2 20 - 4(-4. 9)(-10) 2 ( -4. 9)
t= -20 + √ 2 20 - 4(-4. 9)(-10) 2 ( -4. 9) t = -20 + √ 204 - 9. 8 t =. 6 or t = 3. 5
d. When will ball hit ground? Looking for t? Or h? ? h = 0 when the ball hits ground 2 + t 0 = - 4. 9 20 t + 15 2 0= Ax + Bx+ C
2 + t 0 = - 4. 9 20 t + 15 2 0= Ax + Bx+ C Use quadratic formula t = 4. 68 or t = -. 6 (is this ok? )
When will the ball be at the highest point? ( Max) 2 t + h = - 4. 9 20 t + 15 It will be at the highest point at the vertex. (t, h) So find -b 2 a
vertex. (t, h) So find -b/2 a t value of the Vertex is -20/2 (-4. 9) = 2. 04 So this is the time at max height.
Another method to find vertex Take the x intercepts t = 4. 68 or t = -. 6 4. 68 + -. 6 2 , h
4. 68 + -. 6 2 , h (2. 04, h )
What IS the maximum height? ? Maximum height is at the vertex. If the time at the vertex is 2. 04 how do we get maximum height?
How do we get maximum height? (2. 04, ? ) ( t, h) Substitute 2. 04 in eq. 2 h = - 4. 9 t + 20 t + 15 To get h = 35. 41
I know how to get maximum heigh Just graph the equation. Find th y value of the m
For height look the vertex. To find the time that the ball is in the air, let y = 0, since height = 0 when the ball hits the ground.
Ex 3 Air conditioners Day 0 1 Price 350 340 2 3 320 290 Use Calc. Enter as a list and use quadratic regression. y=- 2 5 x -5 x + 350
Effort It is a solo performance that only you can control. "
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