HURRY PLEASE NEED THIS ASAP The graph below shows a company's profit f(x), in dollars, depending on the price of erasers x, in dollars, sold by the company:

Graph of quadratic function f of x having x intercepts at ordered pairs 0, 0 and 8, 0. The vertex is at 4, 270.

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 = 1 to x = 4, and what does this rate represent? (3 points)

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

Part A i) The x-intercept, represents the points where, f(x) = 0, which are the solution to the function, f(x)

The maximum value represents the vertex

ii) The function is increasing in the interval  -∞ < x ≤ 4 where increased price increases profit

The function is decreasing in the interval  4 ≤ x < ∞, where decreased price decreased profit

Part B:

50.625

Part C:

The constraint of the domain is 0 < x < 8

Step-by-step explanation:

The ordered pairs of the x-intercepts of the quadratic function = (0, 0) and (8, 0)

The vertex = (4, 270)

Part A:

i) The x-intercept represents the solution to the quadratic function f(x) and the values of 'x' for which f(x) = 0

The vertex represents the value of 'x' for which the function f(x) is maximum

maximum value of the function f(x)

ii) The intervals where the function is increasing is to the left of the maximum point, which is -∞ < x ≤ 4

The interval where the function increases represent the relationship between price and profit, such that the profit increases with increase in price to the point where the sale price is $4

The intervals where the function is decreasing is to the right of the maximum point, which is 4 ≤ x < ∞

The interval where the function decreases represent the relationship between price and profit, such that the profit decreases with increase in price from the point where the sale price is $4 and above

Part B

f(x) =a·x² + b·x + c

When x = 0, f(x) = 0

∴ c = 0

When x = 8, f(x) = 0

∴ f(8) =a·8² + b·8 = 64·a + 8·b = 0...(i)

From the vertex, we have; f(4) = a·4² + 4·b = 270...(ii)

Multiplying equation (ii) by 2 gives;

∴  2×(a·4² + 4·b = 270) =  32·a + 8·b = 540...(iii)

Subtracting equation (iii) from equation (i) gives;

64·a + 8·b - (32·a + 8·b) = 0 - 540 = -540

32·a = -540

a = -540/32 = -16.875            

b = From equation (i), we have;

64·a + 8·b = 0

∴ 64 × (-16.875) + 8 × b = 0

b = 64 × (16.875)/8 = 135

∴ f(x) = -16.875·x² + 135·x

At x = 1, f(1) =  -16.875×1² + 135×1 = 118.125

At x = 4, f(4) =  -16.875×4² + 135×4 = 270

The approximate rate of change of the graph, Δf(x)/(Δx) between points x = 1 and x = 4, is given as follows;

Δf(x)/(Δx) = (270 - 118.125)/(4 - 1) = 50.625

The approximate rate of change of the graph from x = 1 to x = 4Δf(x)/(Δx) = 50.625

Part C:

The domain of the function f(x) are the possible values of 'x', the sale price

Given that the sale price is always a positive number, we have that the constraints of the domain is x > 0,

To avoid selling at a loss (negative profit), the value in the domain is constrained to include all values of 'x' for which f(x) is positive, therefore, the domain is constrained to, 0 < x < 8.

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step-by-step explanation:

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