2. Aim: How do we apply the quadratic equation? Do Now: Given a equation: a) Find the coordinates of the turning point b) If y = 0, find the values of x HW: Worksheet

3. The turning point is (3,4). This is the maximum point. Notice that the equation has a negative leading coefficient, then there is a maximum point. To find the values of x when y = 0, we simply replace y by 0 then solve the equation for x. A quadratic equation and the parabola can be applied in many real life situations. Here is the simple example:

4. We can treat the equation as the parabola of the advancing path of a baseball. The maximum point the where the ball reaches its maximum height. The x can be use as the number of seconds and the y can be the height in meter or feet. When y = 0 the x are 1 and 5. That means when time is 0 second the height is 0 meter or feet, when time is 5 seconds the ball comes back to the ground.

5. Find the vertex of

6. Joseph threw a waffle ball out of a window that is four units high. The position of the waffle ball is determined by the parabola y = -x² + 4. At how many feet from the building does the ball hit the ground? The ball lands at the solution of this quadratic equation. There are two solutions. One at 2 and the other at − 2. This picture assumes that Joseph threw the ball to the right so that the waffle balls lands at 2. solution of this quadratic equation

7. A ball is dropped from a height of 36 feet. The quadratic equation d = -t² + 36 provides the distance, d, of the ball, after t seconds. After how many seconds, does the ball hit the ground? The ball hits the ground at d = 0. To find the value of t at this point we must solve this quadratic equation. 0 = −t² + 36 t² = 36 t = 6 (Note: t = -6 is also a solution of this equation. However, only the positive solution is valid since we are measuring seconds.)

8. A ball is dropped from a height of 60 feet. The quadratic equation d = −5t² + 60 provides the distance, d, of the ball, after t seconds. After how many seconds, does the ball hit the ground?