Question

Use the table to answer the question. x y −6 25 −1 0 0 1 A parabola passes through the points provided in the table. The standard form equation for a parabola is y=ax2+bx+c . Write the standard form equation for the parabola.(1 point) The equation for the parabola that passes through the given points is

Answer 1

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:
1. \((-6, 25)\)
2. \((-1, 0)\)
3. \((0, 1)\)

We can use these points to create equations.

### Step 1: Set up the equations

1. For the point \((-6, 25)\):
\[
25 = a(-6)^2 + b(-6) + c \implies 25 = 36a - 6b + c \quad \text{(Equation 1)}
\]

2. For the point \((-1, 0)\):
\[
0 = a(-1)^2 + b(-1) + c \implies 0 = a - b + c \quad \text{(Equation 2)}
\]

3. 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.