1. Pairs of sunglasses sold, x thousands
Properties of Quadratic Functions and Maximum and Minimum Values of Quadratic Functions.
Example: Research shows that the number of sunglasses sold, x, and the selling price, p(x) of of the sunglasses varies according to the following table.
Pairs of sunglasses sold, x thousands
0.7
1.4
1.8
2.1
2.9
3.2
Price, p(x)
30.00
27.50
25.00
22.50
20.00
17.50
What is the relation between pairs sold and price?
Enter the data and perform a linear regression.

2. STAT
2nd Y=

3. STAT
y = – 5x + 34

4. Write the revenue function, R(x), based on the demand function.
R(x) = number of pairs of glasses sold
× price of each pair.
= x × demand function
= x × p(x)
= x × (–5x + 34)

5. Research shows that the cost function is C(x) = 3x + 35
The profit function is P(x) = R(x) – C(x)
P(x) = x × (–5x + 34) – (3x + 35)
= – 5x2 + 34x – 3x – 35
= – 5x2 + 31x – 35

7. Consider the function
y = – 5x2 + 31x

8. Find the coordinates of the vertex:
a) f(x) = (x – 4)(x – 10)
The zeros are 4 and 10
The vertex lies on the line halfway between 4 and 10.
Sub x = 7 into the equation
f(7) = (7 – 4)(7 – 10)
= (3)(– 3)
vertex (7, – 9)
= – 9

9. Find the coordinates of the vertex:
b) f(x) = 2x2 – 10x
= 2x(x – 5)
The zeros are 0 and 5.
The vertex lies on the line halfway between 0 and 5.
Sub x = 2.5 into the equation
f(2.5) = 2(2.5)2 – 10(2.5)
= 12.5 – 25
vertex (2.5, – 12.5)
= – 12.5

10. If the x is the number of items sold in thousands and p(x) is the price of each item, determine the revenue function, R(x), and the maximum revenue.
p(x) = – 2x + 12
R(x) = x(– 2x + 12)
The zeros are 0 and 6.
The vertex lies on the line halfway between 0 and 6.
Sub x = 3 into the equation
R(3) = 3(– 2×3 + 12)
= 3(6)
= 18 (max revenue)

11. The profit function is P(x) = R(x) – C(x)
Given the revenue function and the cost function, determine the (i) profit function and (ii) the value of x that maximizes the profit.
R(x) = – x2 + 18x, C(x) = 8x + 12
The profit function is P(x) = R(x) – C(x)
P(x) = – x2 + 18x – (8x + 12)
P(x) = – x2 + 18x – 8x – 12
P(x) = – x2 + 10x – 12
P(5) = – ×5 – 12
P(x) = – x(x – 10) – 12
P(5) = – – 12
P(5) = 13
halfway between 0 and 10 is 5.

12. 120 – 2x
A farmer has 120 m of fencing to enclose a rectangular pen for his animals with a river on one side. What dimensions will produce a maximum area? x x
river