1.  Smoke Jumpers parachute into locations to suppress forest fires  When they exit the airplane, they are in free fall until their parachutes open  https://www.youtube.com/watch?v=sovMpY2 QH-I

2. If the jumper exits the airplane at a height of 554m, how long will the jumper be in free fall before the parachute opens at 300m? 1. Determine the quadratic relation to model the height, H, of the smoke jumper at time t 2. Determine the length of time that the jumper is in free fall. Notice: the problem contains information about the vertex: he jumps at a height of 554m at time t = 0. Note Also: a = -0.5g = -9.8m/s 2 (acceleration due to gravity)

3.  H = a(t – h) 2 + k  H = a(t – 0) 2 + 554  H = at 2 + 554  H = -0.5(9.8)t 2 + 554  H = -4.9t 2 + 554  Therefore, H = -4.9t 2 + 554 is an equation in standard form that models this relationship.

4. So the jumper is in free fall for about 7.2 seconds.

6.  A coffee shop sells a special blend of coffee for $2.60 per mug. The shop sells about 200 mugs per day. Customer surveys show that for every $0.05 decrease in the price, the shop will sell 10 more mugs per day.  a) Determine the maximum daily revenue from coffee sales and the price per mug for this revenue.  b) Write an equation in both standard form and vertex form to model this problem. Then sketch the graph.

7.  Let x represent the number of $0.05 decreases in price  Revenue = (price)(mugs sold)  Price: 2.60 – 0.05x  Mugs Sold: 200 + 10x  Revenue = (2.60 – 0.05x)(200 + 10x)  We want to find the maximum revenue so we need to vertex  We can start by finding the zeros, and then finding the middle of the zeros to determine the equation of the axis of symmetry

9.  We know the vertex is at x = 16, the r value is:  Revenue = (2.60 – 0.05x)(200 + 10x) r = (2.60 – 0.05(16))(200 + 10(16)) r = (1.80)(360) r = 648 a) Therefore, the maximum daily revenue is $648. The price per mug to maximize revenue is: 2.60 – 0.05x at x = 16, so: 2.60 – 0.05(16) = 1.80 So the coffee shop should sell each mug of coffee for $1.80 to achieve the maximum revenue of $648.

10.  b) Write an equation in both standard form and vertex form to model this problem. Then sketch the graph. Here, the vertex is (16, 648) = (h, k), so we can write:  r = a(x – 16) 2 + 648  We know that when x = 0, r = (2.60 – 0.05x)(200 + 10x), so:  r = (2.60)(200) = 520, so a point is (0, 520) – use to find “a”  r = a(x – 16) 2 + 648  520 = a(0 – 16) 2 + 648  520 = a(– 16) 2 + 648  -128 = 256a  -0.5 = a  So, the equation in vertex form is r = -0.5(x – 16) 2 + 648

11.  To get standard form, expand the equation in vertex form.  r = -0.5(x – 16) 2 + 648  r = -0.5(x 2 – 32x + 256) + 648  r = -0.5x 2 + 16x - 128 + 648  r = -0.5x 2 + 16x + 520 is the equation in standard form.  We can sketch the graph because we know the vertex: (16, 648), x-intercepts (x = 52 & x = -20) and the y-intercept (0, 520):

12.  All quadratic relations can be expressed in vertex form and standard form  Quadratic relations that have zeros can also be expressed in factored form  For any parabola, the value of a is the same in all three forms of the equation of the quadratic relation