Barry owns a music store, He recorded the sales of a newly launched album as part of his biweekly inventory. The table shows the number of weeks, x, since the launch of the album and the number of albums sold during that time period, f(x).


x 0 2 4 6 8

f(x) 0 78 152 222 288


Use the data in the table to create the standard form of the function that models this situation.


f(x)_______ = x2 +_________ x + _________

Respuesta :

Answer:

the standard form of the function that models this situation should be [tex]f(x) = -0.5x^{2} +40x[/tex]

Step-by-step explanation:

Lets visualize the table for x and y values first:

x                y

0                   0  

2                  78

4                  152

6                  222

8                  288

It is clear from the problem, we have to obtain the values of a, b, and c from the equation f(x) = ax² + bx + c.

We need to make use of data in the table to get the values a, b and c. As a, b, and c are three unknowns, we will have to find three equation using the values in data table.

So, if we put x = 0, and y = 0 in f(x) = ax² + bx + c, we can get the value of c.

f(x) = ax² + bx + c

0 = 0 + 0 + c

Lets put some points or pairs to find the values of a and b.

Putting the pair (6, 222) in f(x) = ax² + bx + c

f(x) = ax² + bx + c

222 = a(6)² + b(6) + 0

222 = 36a + 6b          

Dividing the above equation by 6 to further reduce it

37 = 6a + b   (Equation 1)

Putting the pair (8, 288) in f(x) = ax² + bx + c

f(x) = ax² + bx + c

288 = a(8)² + b(8) + 0

288 = 64a + 8b

Dividing the above equation by 8 to further reduce it

36 = 8a + b    (Equation 2)

Subtracting Equation 2 i.e. 36 = 8a + b  from Equation 1 i.e. 37 = 6a + b

1 = -2a

a = -0.5

Putting a = -0.5 in Equation 2

36 = 8a + b

36 = 8(-0.5) + b

36 = -4 + b

b = 40

Hence, the standard form of the function that models this situation should be [tex]f(x) = -0.5x^{2} +40x[/tex]

Keywords: standard form, function

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Answer:

f(x) = -1/2x^2 + 40x + 0

This is the correct answer for Plato and Edmentum users.

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