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