graphs of second degree functions

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.