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To solve the problem, we need to complete the table for different values of [tex]\( x \)[/tex] in the polynomial expression [tex]\( -7x^2 + 32x + 240 \)[/tex]. We will then determine the optimal price that the taco truck should sell its tacos for by identifying the value of [tex]\( x \)[/tex] that maximizes the polynomial expression. Here, [tex]\( x \)[/tex] represents the price in whole dollars.
### Step-by-Step Solution:
1. Understanding the Polynomial Expression:
The polynomial given is:
[tex]\[ f(x) = -7x^2 + 32x + 240 \][/tex]
2. Create the Table for Different Values of [tex]\( x \)[/tex]:
We'll evaluate the polynomial for [tex]\( x \)[/tex] ranging from 0 to 50.
- When [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -7(0^2) + 32(0) + 240 = 240 \][/tex]
- When [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = -7(1^2) + 32(1) + 240 = -7 + 32 + 240 = 265 \][/tex]
- When [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = -7(2^2) + 32(2) + 240 = -28 + 64 + 240 = 276 \][/tex]
- When [tex]\( x = 3 \)[/tex]:
[tex]\[ f(3) = -7(3^2) + 32(3) + 240 = -63 + 96 + 240 = 273 \][/tex]
- When [tex]\( x = 4 \)[/tex]:
[tex]\[ f(4) = -7(4^2) + 32(4) + 240 = -112 + 128 + 240 = 256 \][/tex]
- When [tex]\( x = 5 \)[/tex]:
[tex]\[ f(5)= -7(5^2) + 32(5) + 240 = -175 + 160 + 240 = 225 \][/tex]
- When [tex]\( x = 6 \)[/tex]:
[tex]\[ f(6) = -7(6^2) + 32(6) + 240 = -252 + 192 + 240 = 180 \][/tex]
Continuing this way up to [tex]\( x = 50 \)[/tex], we get the following table:
[tex]\[ \begin{array}{|c|c|} \hline x & f(x) \\ \hline 0 & 240 \\ 1 & 265 \\ 2 & 276 \\ 3 & 273 \\ 4 & 256 \\ 5 & 225 \\ 6 & 180 \\ 7 & 121 \\ 8 & 48 \\ 9 & -39 \\ 10 & -140 \\ 11 & -255 \\ 12 & -384 \\ 13 & -527 \\ 14 & -684 \\ 15 & -855 \\ 16 & -1040 \\ 17 & -1239 \\ 18 & -1452 \\ 19 & -1679 \\ 20 & -1920 \\ 21 & -2175 \\ 22 & -2444 \\ 23 & -2727 \\ 24 & -3024 \\ 25 & -3335 \\ 26 & -3660 \\ 27 & -3999 \\ 28 & -4352 \\ 29 & -4719 \\ 30 & -5100 \\ 31 & -5495 \\ 32 & -5904 \\ 33 & -6327 \\ 34 & -6764 \\ 35 & -7215 \\ 36 & -7680 \\ 37 & -8159 \\ 38 & -8652 \\ 39 & -9159 \\ 40 & -9680 \\ 41 & -10215 \\ 42 & -10764 \\ 43 & -11327 \\ 44 & -11904 \\ 45 & -12495 \\ 46 & -13100 \\ 47 & -13719 \\ 48 & -14352 \\ 49 & -14999 \\ 50 & -15660 \\ \hline \end{array} \][/tex]
3. Determine the Optimal Price:
We observe that the value of [tex]\( f(x) \)[/tex] is maximized when [tex]\( x = 2 \)[/tex]. At this price point, [tex]\( f(x) \)[/tex] yields the highest value, which is [tex]\( 276 \)[/tex].
### Conclusion:
The optimal price that the taco truck should sell its tacos for is [tex]\( \boxed{2} \)[/tex] dollars. This price maximizes the polynomial expression, corresponding to a value of 276.
### Step-by-Step Solution:
1. Understanding the Polynomial Expression:
The polynomial given is:
[tex]\[ f(x) = -7x^2 + 32x + 240 \][/tex]
2. Create the Table for Different Values of [tex]\( x \)[/tex]:
We'll evaluate the polynomial for [tex]\( x \)[/tex] ranging from 0 to 50.
- When [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -7(0^2) + 32(0) + 240 = 240 \][/tex]
- When [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = -7(1^2) + 32(1) + 240 = -7 + 32 + 240 = 265 \][/tex]
- When [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = -7(2^2) + 32(2) + 240 = -28 + 64 + 240 = 276 \][/tex]
- When [tex]\( x = 3 \)[/tex]:
[tex]\[ f(3) = -7(3^2) + 32(3) + 240 = -63 + 96 + 240 = 273 \][/tex]
- When [tex]\( x = 4 \)[/tex]:
[tex]\[ f(4) = -7(4^2) + 32(4) + 240 = -112 + 128 + 240 = 256 \][/tex]
- When [tex]\( x = 5 \)[/tex]:
[tex]\[ f(5)= -7(5^2) + 32(5) + 240 = -175 + 160 + 240 = 225 \][/tex]
- When [tex]\( x = 6 \)[/tex]:
[tex]\[ f(6) = -7(6^2) + 32(6) + 240 = -252 + 192 + 240 = 180 \][/tex]
Continuing this way up to [tex]\( x = 50 \)[/tex], we get the following table:
[tex]\[ \begin{array}{|c|c|} \hline x & f(x) \\ \hline 0 & 240 \\ 1 & 265 \\ 2 & 276 \\ 3 & 273 \\ 4 & 256 \\ 5 & 225 \\ 6 & 180 \\ 7 & 121 \\ 8 & 48 \\ 9 & -39 \\ 10 & -140 \\ 11 & -255 \\ 12 & -384 \\ 13 & -527 \\ 14 & -684 \\ 15 & -855 \\ 16 & -1040 \\ 17 & -1239 \\ 18 & -1452 \\ 19 & -1679 \\ 20 & -1920 \\ 21 & -2175 \\ 22 & -2444 \\ 23 & -2727 \\ 24 & -3024 \\ 25 & -3335 \\ 26 & -3660 \\ 27 & -3999 \\ 28 & -4352 \\ 29 & -4719 \\ 30 & -5100 \\ 31 & -5495 \\ 32 & -5904 \\ 33 & -6327 \\ 34 & -6764 \\ 35 & -7215 \\ 36 & -7680 \\ 37 & -8159 \\ 38 & -8652 \\ 39 & -9159 \\ 40 & -9680 \\ 41 & -10215 \\ 42 & -10764 \\ 43 & -11327 \\ 44 & -11904 \\ 45 & -12495 \\ 46 & -13100 \\ 47 & -13719 \\ 48 & -14352 \\ 49 & -14999 \\ 50 & -15660 \\ \hline \end{array} \][/tex]
3. Determine the Optimal Price:
We observe that the value of [tex]\( f(x) \)[/tex] is maximized when [tex]\( x = 2 \)[/tex]. At this price point, [tex]\( f(x) \)[/tex] yields the highest value, which is [tex]\( 276 \)[/tex].
### Conclusion:
The optimal price that the taco truck should sell its tacos for is [tex]\( \boxed{2} \)[/tex] dollars. This price maximizes the polynomial expression, corresponding to a value of 276.
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