Question

QUADRATIC FUNCTIONS: The profit (in hundreds of dollars) that a corporation receives depends on the amount (in hundreds of dollars) the company spends on marketing according to the model 130+10X−0.5x2130+10X−0.5x2. What expenditure for advertising yields a maximum profit? What is the mathematical name of this point?

PROBLEM 4: POLYNOMIAL FUNCTIONS: Let f(x)=4x5−8x4−5x3+10x2+x−1f(x)=4x5−8x4−5x3+10x2+x−1. The graph is presented below:
Describe f (x) in terms of
Degree of polynomial
Main coefficient
Final behavior
Maximum number of zeros
Maximum number of exchange points (relative maximums and minimums)

Answer 1

Explanation:

The profit (in hundreds of dollars) that a corporation receives is given by the quadratic function:

P(x) = 130 + 10x - 0.5x^2

where x is the amount spent on marketing (in hundreds of dollars).

To find the expenditure for advertising that yields a maximum profit, we need to find the vertex of the parabola. The vertex occurs at:

x = -b/(2a) = -10/(2*(-0.5)) = 10

Substituting x = 10 back into the equation for P(x), we get:

P(10) = 130 + 10(10) - 0.5(10)^2 = 180

Therefore, an expenditure of $1000 for advertising yields a maximum profit of $18000.

The mathematical name of the point is the vertex of the parabola.

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For the polynomial function:

f(x) = 4x^5 - 8x^4 - 5x^3 + 10x^2 + x - 1

Degree of polynomial: 5

Main coefficient: 4 (the leading coefficient)

Final behavior: As x approaches positive or negative infinity, f(x) also approaches positive infinity (since the leading term has a positive coefficient and has the highest degree).

Maximum number of zeros: 5 (since it is a fifth-degree polynomial)

Maximum number of exchange points: 4 (since there are 4 relative extrema, either maximum or minimum points)