Question

Remington wanted to write the parabolic equation with the three points (0,3) , (−3,2) , and (3,4) . Is this possible? What does the result mean?(1 point)
Responses

This is not possible. Since a=0, the result is a linear equation, y=13x+3.
This is not possible. Since a equals 0 , the result is a linear equation, y equals Start Fraction 1 over 3 End Fraction x plus 3 .

This is possible. The parabolic equation is y=23x2+73x+3.
This is possible. The parabolic equation is y equals Start Fraction 2 over 3 End Fraction x squared plus Start Fraction 7 over 3 End Fraction x plus 3 .

This is not possible. Since a=0, only higher degree equations can be written.
This is not possible. Since a equals 0 , only higher degree equations can be written.

This is possible. The parabolic equation is y=13x2+3.

Answers

There are no human answers yet.
Answered by GPT-4o mini
To determine whether it is possible to fit a parabolic equation (in the form \(y = ax^2 + bx + c\)) to three given points \((0,3)\), \((-3,2)\), and \((3,4)\), we can substitute these points into the equation to create a system of equations.

1. For the point \((0,3)\):
\[
3 = a(0)^2 + b(0) + c \implies c = 3
\]

2. For the point \((-3,2)\):
\[
2 = a(-3)^2 + b(-3) + c \implies 2 = 9a - 3b + 3
\]
Simplifying gives:
\[
9a - 3b = -1 \quad \text{(Equation 1)}
\]

3. For the point \((3,4)\):
\[
4 = a(3)^2 + b(3) + c \implies 4 = 9a + 3b + 3
\]
Simplifying gives:
\[
9a + 3b = 1 \quad \text{(Equation 2)}
\]

Now we have two equations:
1. \(9a - 3b = -1\) (Equation 1)
2. \(9a + 3b = 1\) (Equation 2)

We can add these two equations to eliminate \(b\):
\[
(9a - 3b) + (9a + 3b) = -1 + 1
\]
This gives:
\[
18a = 0 \implies a = 0
\]

Substituting \(a = 0\) back into either equation will provide \(b\). Using Equation 2:
\[
9(0) + 3b = 1 \implies 3b = 1 \implies b = \frac{1}{3}
\]

Thus:
- \(a = 0\)
- \(b = \frac{1}{3}\)
- \(c = 3\)

The resulting equation is:
\[
y = 0x^2 + \frac{1}{3}x + 3 \implies y = \frac{1}{3}x + 3
\]

This is a linear equation, meaning that, while we can fit these points with a linear function, we cannot fit them to a quadratic function (parabola). Therefore, the concluding statement is:

**This is not possible. Since \(a = 0\), the result is a linear equation, \(y = \frac{1}{3}x + 3\).**

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