The graph below shows a company's profit f(x), in dollars, depending on the price of pens x, in dollars, sold by the company: Graph of quadratic function f of x having x intercepts at ordered pairs 0, 0 and 6, 0. The vertex is at 3, 120.

Part A: What do the x-intercepts and maximum value of the graph represent? What are the intervals where the function is increasing and decreasing, and what do they represent about the sale and profit? (4 points)

Part B: What is an approximate average rate of change of the graph from x = 3 to x = 5, and what does this rate represent? (3 points)

Part C: Describe the constraints of the domain. (3 points)

a) At any x-intercept, f(x) = 0. In the context of the problem, this means that the company makes 0 profit when they price the pens at 0 and at 6.

The interval where the graph is increasing is 0<x<3, and decreasing at 3<x<6 Which means from $0 to $3, the total profit increases, and as the pens are priced more and more above $3, the total profit decreases until it gets to 0 at $6 (because nobody wants to buy them).

b) I dont know your graph at x = 5 but x = 3 is the maximum so the rate has to be negative. The negative rate means that the company is losing profit as they charge more and more above $3. T

c) The domain is limited to 0≤x≤6 because if they charge any more or any less they would not make negative money.

answer: c, the green line

step-by-step explanation:

there is no black line, eliminate b.

simple interest implies periodic payments, but the red line shows no payments, eliminate d.

simple interest has equal payments. if paid say every day, the graph is about a straight line (actually a stairstep).

the yellow line is not straight, eliminate a.

check: the green line is straight, answer is c.

simple interest with positive interest rate gives a straight line sloping up to the right. green line:

g(t) = $100 + ($200-$100) ÷ (20 years) × (t years)

= $100 + (4%/year) × ($100)(t years)

incidentally, the yellow line curving upward represents continuously compounded interest

y = $100 × 2^(t years / (14 assuming the yellow line hits $200 at 14 years. 2^0 is 1, so y(0) = $100; 2^(14/14) is 2, so y(14) = $200.