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 possible. The parabolic equation is
y
=
1
3
x
2
+
3
.
This is possible. The parabolic equation is y equals Start Fraction 1 over 3 End Fraction x squared plus 3 .
This is possible. The parabolic equation is
y
=
2
3
x
2
+
7
3
x
+
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 not possible. Since
a
=
0
, the result is a linear equation,
y
=
1
3
x
+
3
.
(
0
,
3
)
,
(
−
3
,
2
)
, and
(
3
,
4
)
. Is this possible? What does the result mean?(1 point)
Responses
This is possible. The parabolic equation is
y
=
1
3
x
2
+
3
.
This is possible. The parabolic equation is y equals Start Fraction 1 over 3 End Fraction x squared plus 3 .
This is possible. The parabolic equation is
y
=
2
3
x
2
+
7
3
x
+
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 not possible. Since
a
=
0
, the result is a linear equation,
y
=
1
3
x
+
3
.
Answers
GPT-4o mini
To determine whether a parabolic equation can be written through the given three points, we need to check if these points are compatible and lead to a unique quadratic function. The general form of a parabolic equation is given by:
\[ y = ax^2 + bx + c \]
Given the points \( (0, 3) \), \( (-3, 2) \), and \( (3, 4) \), we can derive the coefficients \( a \), \( b \), and \( c \).
1. For the point \( (0, 3) \):
\[ 3 = a(0)^2 + b(0) + c \]
So, \( c = 3 \).
2. For the point \( (-3, 2) \):
\[ 2 = a(-3)^2 + b(-3) + 3 \]
\[ 2 = 9a - 3b + 3 \]
Simplifying, we find:
\[ 9a - 3b = -1 \quad \text{(1)} \]
3. For the point \( (3, 4) \):
\[ 4 = a(3)^2 + b(3) + 3 \]
\[ 4 = 9a + 3b + 3 \]
Simplifying, we find:
\[ 9a + 3b = 1 \quad \text{(2)} \]
Now we have a system of equations (1) and (2):
From equation (1):
\[ 9a - 3b = -1 \]
From equation (2):
\[ 9a + 3b = 1 \]
Next, we can solve these equations. Adding both equations:
\[
(9a - 3b) + (9a + 3b) = -1 + 1
\]
\[
18a = 0 \implies a = 0
\]
Plugging \( a = 0 \) back into either equation will yield \( b \). Let's use (2):
\[
9(0) + 3b = 1 \implies 3b = 1 \implies b = \frac{1}{3}
\]
So we can then express the equation:
\[
y = 0 \cdot x^2 + \frac{1}{3}x + 3
\]
This simplifies to a linear equation:
\[
y = \frac{1}{3}x + 3
\]
### Conclusion
Since we derived \( a = 0 \), that means the equation is indeed linear, not quadratic.
Thus, the correct answer is:
**This is not possible. Since \( a = 0 \), only higher degree equations can be written.**
\[ y = ax^2 + bx + c \]
Given the points \( (0, 3) \), \( (-3, 2) \), and \( (3, 4) \), we can derive the coefficients \( a \), \( b \), and \( c \).
1. For the point \( (0, 3) \):
\[ 3 = a(0)^2 + b(0) + c \]
So, \( c = 3 \).
2. For the point \( (-3, 2) \):
\[ 2 = a(-3)^2 + b(-3) + 3 \]
\[ 2 = 9a - 3b + 3 \]
Simplifying, we find:
\[ 9a - 3b = -1 \quad \text{(1)} \]
3. For the point \( (3, 4) \):
\[ 4 = a(3)^2 + b(3) + 3 \]
\[ 4 = 9a + 3b + 3 \]
Simplifying, we find:
\[ 9a + 3b = 1 \quad \text{(2)} \]
Now we have a system of equations (1) and (2):
From equation (1):
\[ 9a - 3b = -1 \]
From equation (2):
\[ 9a + 3b = 1 \]
Next, we can solve these equations. Adding both equations:
\[
(9a - 3b) + (9a + 3b) = -1 + 1
\]
\[
18a = 0 \implies a = 0
\]
Plugging \( a = 0 \) back into either equation will yield \( b \). Let's use (2):
\[
9(0) + 3b = 1 \implies 3b = 1 \implies b = \frac{1}{3}
\]
So we can then express the equation:
\[
y = 0 \cdot x^2 + \frac{1}{3}x + 3
\]
This simplifies to a linear equation:
\[
y = \frac{1}{3}x + 3
\]
### Conclusion
Since we derived \( a = 0 \), that means the equation is indeed linear, not quadratic.
Thus, the correct answer is:
**This is not possible. Since \( a = 0 \), only higher degree equations can be written.**