Quadratic Function model

1. Quadratic Function model

2. Quadratic Function

3. Quadratic Formula • For all a,b, and c that are real numbers and a is not equal to zero

4. Standard Form of a Quadratic Function • Graph is a parabola • Axis is the vertical line x = h • Vertex is (h,k) • a>0 graph opens upward • a<0 graph opens downward

5. Find Vertex • x coordinate is • y coordinate is

6. The general function • The general function that approximate the height h of a projectile on earth after t second • h(t) = - (1 / 2) g t 2 + Vo t + Ho • For a quadratic function of the form  • h(t) = a t 2 + b t + cand with a negative, the maximum of h occur at t = - b / 2a (vertex of the graph of h)

7. Example: The formula h (t) = -16 t 2 + 32 t + 80 gives the height h above ground, in feet, of a Football thrown, at t = 0, straight upward from the top of an 80 feet building. a - What is the highest point reached by the Football?  b-How long does it take the Football to reach its highest point? 

8. c - After how many seconds does the Football hit the ground?  d - For how many seconds is the height of the Football higher than 90 feet?

9.  Maximum height of the Football • The height h given above is a quadratic function. The graph of h as a function of time t gives a parabolic shape and the maximum height h occur at the vertex of the parabola. For a quadratic function of the form h = a t 2 + b t + c, the vertex is located at t = - b / 2a. Hence for h given above the vertex is at t  • t = -32 / 2(-16) = 1 second.  • 1 second after the Football was thrown, it reaches its highest point (maximum value of h) which is given by h = -16 (1) 2 + 32 (1) + 80 = 96 feet

10.   How long does it take the Football to reach its highest point? •  It takes 1 second for the ball to reach it highest point

11. After how many seconds does the ball hit the ground?  • At the ground h = 0, hence the solution of the equation h = 0 gives the time t at which the ball hits the ground. -16 t 2 + 32 t + 80 = 0  • The above quadratic equation has two solutions one is negative and the second one is positive and approximately equal to 3.5 seconds. So it takes 3.5 seconds for the ball to hit the ground after it has been thrown upward

12. For how many seconds is the height of the ball higher than 90 feet? • The ball is higher than 90 feet for all values of t satisfying the inequality h > 90. Hence -16 t 2 + 32 t + 80 > 90 • The above inequality is satisfied for 0.4 < t < 1.6 (seconds)  • The height of the ball is higher than 90 feet for 1.6 - 0.4 = 1.2 seconds.  • The graph below is that of h in terms of t and clearly shows that h is greater than 90 feet for t between 0.4 and 1.6 seconds. 

13. MODELLING A FOOTBALL THROW vertex(5m,5m) y = a(x-5)2 + 5 How can we find ‘a’ x-int: (11,0)

14. How to model a quadratic relation when you have a vertex and one other point • We know the quadratic relation is of the form:    y = a(x-5)2 + 5   - because the vertex is (5,5) • We can use the point (11,0) and substitute those values for x and y • 0 = a(11-5)2 + 5 • 0 = 36a + 5 • -5 = 36a • a = -5/36 • y = (-5/36)(x-5)2 + 5

15. References • https://www.google.ca/search?q=quadratic+function+football+model+powerpoint • http://www.slideshare.net/astarre33/quad-fcn • https://www.google.ca/search?q=football+kicked+image