2.4 graphs of second degree functions t
- 1. Example C. Graph y = x2 – 4x – 12 Vertex: set x = = 2 then y = 22 – 4*2 – 12 = –16, –(–4) 2(1) so v = (2, –16). (2, –16) (0, –12) (4, –12) Second Degree Functions Following is an example of maximization via the vertex. Another point: Set x = 0 then y = –12 or (0, –12) It's reflection across the mid–line is (4, –12) Set y = 0 and get x–int: x2 – 4x – 12 = 0 (x + 2)(x – 6) = 0 x = –2, x = 6
- 2. Example D. The "Crazy Chicken" can sell 120 whole roasted chicken for $8 in one day. For every $0.50 increases in the price, 4 less chickens are sold. Find the price that will maximize the revenue (the sale). Second Degree Functions Set x to be the number of $0.50 increments above the base price of $8.00. So the price is (8 + 0.5x) or (8 + x/2). The number of chicken sold at this price is (120 – 4x). Hence the revenue, depending on the number of times of $0.50 price hikes x, is R(x) = (8 + x/2)(120 – 4x) = 960 + 28x – 2x2. This a 2nd degree equation whose graph is a parabola that opens downward. The vertex is the highest point on the graph where R is the largest.
- 3. Second Degree Functions x RThe vertex of this parabola is at x = = 7. So R(7) = $1058 gives the maximum revenue. –(28) 2(–2) R = 960 + 28x – 2x2. v = (7, 1058) More precisely, raise the price x = 7 times to 8 + 7(0.50) = 11.50 per chicken,chicken, and we can sell 120 – 4(7) = 92 chickens per day with the revenue of 11.5(92) = 1058. Any inverse square law in science is 2nd degree. Laws related to distance or area in mathematics are 2nd degree. That’s why 2nd equations are important.