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QuestionFill in missing values for the function mapping: 1 to 6, 4 to 16, 7 to 1283\frac{128}{3}. Find the function's equation.

Studdy Solution

STEP 1

Assumptions1. The function maps the integers from1 to7 to certain values. . The given mappings are1 to6,4 to16, and7 to128/3.
3. The function follows a certain rule to transform each input (x) to its output (y).

STEP 2

Let's observe the given mappings. We notice that the output increases as the input increases, suggesting that the function might be a polynomial function.

STEP 3

Let's assume that the function is a polynomial of degree n. Since we have3 points, we can assume that the polynomial is of degree2 (a quadratic function), because a polynomial of degree n can be determined by (n+1) points.

STEP 4

Let's write the general form of a quadratic function.
y=ax2+bx+cy = ax^2 + bx + c

STEP 5

We can form a system of equations using the given mappings (1,), (4,16), and (7,128/3) and the general form of the quadratic function.
From (1,):
=a+b+c(1) = a + b + c \tag{1}From (4,16):
16=16a+4b+c(2)16 =16a +4b + c \tag{2}From (7,128/3):
1283=49a+7b+c(3)\frac{128}{3} =49a +7b + c \tag{3}

STEP 6

We can solve this system of equations to find the values of a, b, and c. Let's first subtract equation (1) from equation (2) and equation (3).
From (2)-(1):
16−6=16a−a+4b−b16 -6 =16a - a +4b - b10=15a+3b(4)10 =15a +3b \tag{4}From (3)-(1):
1283−6=49a−a+b−b\frac{128}{3} -6 =49a - a +b - b1103=48a+6b(5)\frac{110}{3} =48a +6b \tag{5}

STEP 7

We can simplify equation (5) by dividing by3.
1109=16a+2b(6)\frac{110}{9} =16a +2b \tag{6}

STEP 8

Now we can subtract equation (4) from equation (6) to find the value of a.
110−10=16a−15a\frac{110}{} -10 =16a -15a20=a\frac{20}{} = a

STEP 9

Substitute a =20/9 into equation (4) to find the value of b.
=15⋅209+3b =15 \cdot \frac{20}{9} +3bb=309=3b = \frac{30}{9} = \frac{}{3}

STEP 10

Substitute a =20/9 and b =10/3 into equation () to find the value of c.
6=209+103+c6 = \frac{20}{9} + \frac{10}{3} + cc=6−209−103=2c =6 - \frac{20}{9} - \frac{10}{3} =2

STEP 11

Now that we have the values of a, b, and c, we can write the equation of the function.
y=209x+103x+y = \frac{20}{9}x^ + \frac{10}{3}x +

STEP 12

Now we can use this equation to fill in the missing values in the table. For x =2,,5, and6, we substitute these values into the equation to find the corresponding y values.
For x =2y=209(2)2+10(2)+2=52y = \frac{20}{9}(2)^2 + \frac{10}{}(2) +2 = \frac{52}{}For x =y=209()2+10()+2=10y = \frac{20}{9}()^2 + \frac{10}{}() +2 =10For x =5y=209(5)2+10(5)+2=130y = \frac{20}{9}(5)^2 + \frac{10}{}(5) +2 = \frac{130}{}For x =6y=209(6)2+10(6)+2=28y = \frac{20}{9}(6)^2 + \frac{10}{}(6) +2 =28

STEP 13

The completed table of values for the function is\begin{tabular}{|c|c|c|c|c|c|c|} \hline &2 &3 & &5 &6 &7 \\ \hline6 & 523\frac{52}{3} &10 &16 & 1303\frac{130}{3} &28 & 1283\frac{128}{3} \\ \hline\end{tabular}
The equation of the function isy=209x2+103x+2y = \frac{20}{9}x^2 + \frac{10}{3}x +2

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