Ashton works as a marketing manager for a consumer products firm. His total earnings, A, in x months is given by the function A(x) = 2,400x + 40x2. His total savings, S, in x months is given by the function S(x) = 250x + 500.
Write a function that represents Ashton's total expenses in months.
How much would Ashton's expenses be over 3 years?

Respuesta :

If A(x) is his total earnings and S(x) is his total savings, his total expenses is the difference between his total earnings and total savings. 

A(x) = 2400x +40x^2
S(x) = 250x + 500

let E(x) be the total expenses:

E(x) = A(x) - S(x)
E(x) = 2400x + 40x^2 - (250x +500)
E(x) = 40x^2 + 2150x - 500

Taking into account the operations between polynomial functions, the funtion E(x)= 40x² +2,150x - 500, where x represents the number of months, represents Ashton's total expenses in months and Ashton's expenses over 3 years would be 128,740.

Subtraction of polynomials

Polynomials are algebraic expressions consisting of variables, constants, and exponents that are combined using mathematical operations such as addition, subtraction, and multiplication.

To subtract two polynomials, you must subtract the terms of the polynomials that are similar. That is, the subtraction of polynomials consists of subtracting the terms that have the same variables raised to the same powers (same exponents).

Function that represents Ashton's total expenses in months

Ashton works as a marketing manager for a consumer products firm.

His total earnings, A, in x months is given by the function A(x)=2,400x+40x².

His total savings, S, in x months is given by the function S(x)= 250x+500.

So, his total expenses E(x) is the difference between his total earnings and total savings:

E(x) = A(x) - S(x)

Replacing the corresponding expressions:

E(x)= 2,400x+40x² - (250x+500)

Solving the subtraction:

E(x)= 40x² +2,400x - 250x - 500

E(x)= 40x² +2,150x - 500 where x represents the number of months.

Ashton's expenses over 3 years

Since 1 year represents 12 months, 3 years contains 36 months. Then, substituting in the function corresponding to the expense E(x) and solving, you obtain:

E(36)= 40×(36)² +2,150×36 - 500

E(36)= 128,740

Summary

Finally, the funtion E(x)= 40x² +2,150x - 500, where x represents the number of months, represents Ashton's total expenses in months and Ashton's expenses over 3 years would be 128,740.

Learn more about subtraction of polynomials:

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