Question

PLEASE HELP ASAP!! FOR 50 POINTS

All that is needed to do is to prove that the data satisfies a quadratic function​

Answer 1

• The second differences of the given data are all equal and therefore, it satisfies a quadratic function.

Task :

• To prove that the given data satisfies a quadratic function.

Solution :

We can prove that the given data satisfies a quadratic function easily by finding out the second differences and if all of them are equal,it means the data satisfies a quadratic function.

In order to find second differences, we'll firstly find the first differences by deducting the first value of y from the preceding value of y and so on and further, we'll find the second differences by following the same Pattern.

First differences :

• 80 - 0 = 80
• 128 - 80 = 48
• 144 - 128 = 16
• 128 - 144 = -16

Second differences :

• 48 - 80 = -32
• 16 - 48 = -32
• - 16 - 16 = -32

Since, the second differences are same ,thus the given data must satisfy a quadratic function.

Hence ,proved ✓.

Moreover ,

If we are asked to find out the quadratic function,

we can simply substitute the value of x and y in the quadratic formula that is given by :

• 
  `ax {}^(2) + by \: + c \: = y`

when x = 0 , y = 0,

• 
  `a(0 {}^(2) ) + b(0) + c = 0 \\ c = 0`

Since, the value of c = 0 ,thus, we'll leave it out from the equation in the further work.

when x = 1 and y = 80,

• 
  `a(1 {}^(2) ) + b(1) = 80 \\ a + b = 80.......(1)`

when x = 2 and y = 3

• 
  `a(2 {}^(2)) + b(2) = 128 \\4a + 2b = 128......(2)`

Now, to find the value of a and b , we'll multiply equation (1) by 4 and subtract equation (2) from it

• 
  `4a +4b = 320 \\ - 4a + 2b = 128 \\ \underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ 2b = 192 \\ \underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ b = 96 \\ \underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }`

Plug in the value of b in equation (1),

• 
  `a + 96 = 80 \\ a = 80 - 96 \\ a = - 16`

Thus, a = -16 and b = 96 and therefore,our quadratic function is :

• 
  `y = - 16x {}^(2) + 96x`

In order to check if it is right for rest of the data , we'll plug in the values for the set (3,144),(4,128)

• 
  `144 = - 16(3) {}^(2) + 96 * 3 \\ 144 = - 16 * 9 + 288 \\ 144 = 144`
• 
  `128 = - 16(4) {}^(2) + 96 * 4 \\ 128 = - 256 + 384 \\ 128 = 128`

Since ,LHS = RHS,hence ,the given data works perfectly for the quadratic function ( -16x^2 + 96x = y)

Answer 2

To determine that the data satisfies a quadratic function, we need to work out the second differences between the y-values. If the second differences are the same, this proves that the function is quadratic.

Explanation:

To determine the type of function that a given set of data satisfies, we need to calculate the differences between the y-values until those differences become consistent.

This degree of consistency in the differences corresponds to the degree of the polynomial function. For instance

• If the first differences are the same, the function is linear.
• If the second differences are the same, the function is quadratic.
• If the third differences are the same, the function is cubic.

Therefore, to prove that the given data satisfies a quadratic function, we need to determine the second differences.

First, calculate the first differences between the y-values:

`0 \underset{+80}{\longrightarrow} 80\underset{+48}{\longrightarrow}128\underset{+16}{\longrightarrow}144\underset{-16}{\longrightarrow}128`

Now, determine the second differences by calculating the differences between the first differences:

`80 \underset{-32}{\longrightarrow} 48\underset{-32}{\longrightarrow} 16\underset{-32}{\longrightarrow} -16`

As the second differences are the same (-32), this proves that the data satisfies a quadratic function.

`\hrulefill`

Finding the function

As the function is quadratic, it will contain an x² term.

The coefficient of x² is always half the second difference. Therefore, the coefficient of x² = -16.

Determine the operation that takes -16x² to y:

`\begin{array}c\cline{1-6}x&0&1&2&3&4\\\cline{1-6}-16x^2&-0&-16&-64&-144&-256\\\cline{1-6}\sf Operation&+0&+96&+192&+288&+384\\\cline{1-6}y&0&80&128&144&128\\\cline{1-6}\end{array}`

As the operation is not constant, work out the differences between the operations:

`0 \underset{+96}{\longrightarrow} 96\underset{+96}{\longrightarrow} 192\underset{+96}{\longrightarrow} 288\underset{+96}{\longrightarrow} 384`

As the differences are the same, the second operation in the function is +96x. Write out the x-values with both operations:

`\begin{array}c\cline{1-6}x&0&1&2&3&4\\\cline{1-6}-16x^2&-0&-16&-64&-144&-256\\\cline{1-6}+96x&+0&+96&+192&+288&+384\\\cline{1-6}y&0&80&128&144&128\\\cline{1-6}\end{array}`

As -16x² + 96x takes x to y, the equation of the function is:

`y=-16x^2+96x`