To find the standard form equation of the parabola \( y = ax^2 + bx + c \) that passes through the given points, we can set up a system of equations using the points provided in the table.
The points from the table are:
- \((-6, 25)\)
- \((-1, 0)\)
- \((0, 1)\)
We can use these points to create equations.
Step 1: Set up the equations
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For the point \((-6, 25)\): \[ 25 = a(-6)^2 + b(-6) + c \implies 25 = 36a - 6b + c \quad \text{(Equation 1)} \]
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For the point \((-1, 0)\): \[ 0 = a(-1)^2 + b(-1) + c \implies 0 = a - b + c \quad \text{(Equation 2)} \]
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For the point \((0, 1)\): \[ 1 = a(0)^2 + b(0) + c \implies 1 = c \quad \text{(Equation 3)} \]
Step 2: Substitute Equation 3 into Equations 1 and 2
From Equation 3, we have \( c = 1 \).
Substituting \( c = 1 \) into Equation 1: \[ 25 = 36a - 6b + 1 \] \[ 24 = 36a - 6b \quad \text{(Equation 4)} \]
Substituting \( c = 1 \) into Equation 2: \[ 0 = a - b + 1 \] \[ a - b = -1 \quad \text{(Equation 5)} \]
Step 3: Solve the system of equations
Now we can solve Equations 4 and 5.
From Equation 5: \[ a = b - 1 \quad \text{(Equation 6)} \]
Substituting Equation 6 into Equation 4: \[ 24 = 36(b - 1) - 6b \] \[ 24 = 36b - 36 - 6b \] \[ 24 + 36 = 30b \] \[ 60 = 30b \implies b = 2 \]
Using \( b = 2 \) in Equation 6 to find \( a \): \[ a = 2 - 1 = 1 \]
Now we have:
- \( a = 1 \)
- \( b = 2 \)
- \( c = 1 \)
Final Step: Write the standard form equation
Thus, the standard form equation for the parabola is: \[ \boxed{y = x^2 + 2x + 1} \]
You can also rewrite this equation as: \[ y = (x + 1)^2 \] which confirms that it passes through the given points.
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