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Use quadratic regression to find a

function that fits the following
points.
(-1,-15), (1,-7), (6,-22)
y = [ ? ]x² + [ ]x + []

User Maestro
by
7.1k points

1 Answer

2 votes

Answer:

To fit a quadratic function to the given points, we'll use the general form of a quadratic equation:

\[ y = ax^2 + bx + c \]

where \( a \), \( b \), and \( c \) are the coefficients we need to find.

To find these coefficients, we can set up a system of equations using the three given points and solve for \( a \), \( b \), and \( c \).

Given points: (-1, -15), (1, -7), and (6, -22)

1. For point (-1, -15):

\[ -15 = a(-1)^2 + b(-1) + c \]

\[ -15 = a - b + c \]

2. For point (1, -7):

\[ -7 = a(1)^2 + b(1) + c \]

\[ -7 = a + b + c \]

3. For point (6, -22):

\[ -22 = a(6)^2 + b(6) + c \]

\[ -22 = 36a + 6b + c \]

Now, we have a system of three equations:

1. \( -15 = a - b + c \)

2. \( -7 = a + b + c \)

3. \( -22 = 36a + 6b + c \)

Let's solve this system to find the values of \( a \), \( b \), and \( c \).

Subtracting the second equation from the first equation to eliminate \( b \):

\( (-15) - (-7) = a - b + c - a - b - c \)

\( -8 = -2b \)

\( b = 4 \)

Now, we can substitute the value of \( b \) back into the second equation to find the value of \( a \):

\( -7 = a + 4 + c \)

\( -7 - 4 = a + c \)

\( a = -11 - c \)

Now, we can substitute the values of \( a \) and \( b \) into the third equation to solve for \( c \):

\( -22 = 36(-11 - c) + 6(4) + c \)

\( -22 = -396 - 36c + 24 + c \)

\( -22 = -372 - 35c \)

\( -35c = -22 + 372 \)

\( -35c = 350 \)

\( c = \frac{-350}{-35} \)

\( c = 10 \)

Now that we have the value of \( c \), we can find the value of \( a \):

\( a = -11 - c \)

\( a = -11 - 10 \)

\( a = -21 \)

So, the quadratic function that fits the given points is:

\[ y = -21x^2 + 4x + 10 \]

User Omar AMEZOUG
by
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