The figure shows the graph of y equals negative ax squared plus bx plus c. Suppose that the points left parenthesis negative h comma y 0 right parenthesis comma left parenthesis 0 comma y 1 right parenthesis comma and left parenthesis h comma y 2 right parenthesis are on the graph. The area enclosed by the​ parabola, the​ x-axis, and the lines x equals negative h and x equals h may be given by the formula below. Find the area enclosed by f left parenthesis x right parenthesis equals negative 6 x squared plus 7 comma the​ x-axis, and the lines xequalsminus1 and xequals1.
AreaequalsStartFraction h Over 3 EndFraction left parenthesis y 0 plus 4 y 1 plus y 2 right parenthesis
x
y
left parenthesis negative h comma y 0 right parenthesisleft parenthesis 0 comma y 1 right parenthesisleft parenthesis h comma y 2 right parenthesis

A coordinate system has a horizontal x-axis and a vertical y-axis. A parabola opens downward from the plotted point (0, y 1) and passes through the plotted points (negative h, y 0) and (h, y 2).
Question content area bottom
Part 1
The area is

enter your response here.

We are given the equation of the parabola:

\[
f(x) = -6x^2 + 7
\]

We need to calculate the area enclosed by the parabola, the x-axis, and the vertical lines \( x = -1 \) and \( x = 1 \). The formula for the area is:

\[
\text{Area} = \frac{h}{3} \left( y_0 + 4y_1 + y_2 \right)
\]

where:
- \( h = 1 \) (since the limits of integration are \( x = -1 \) and \( x = 1 \)),
- \( y_0 \) is the value of \( f(x) \) at \( x = -h \),
- \( y_1 \) is the value of \( f(x) \) at \( x = 0 \),
- \( y_2 \) is the value of \( f(x) \) at \( x = h \).

### Step 1: Calculate \( y_0 \), \( y_1 \), and \( y_2 \)
1. **At \( x = -1 \) (i.e., \( y_0 \)):**
   \[
   y_0 = f(-1) = -6(-1)^2 + 7 = -6 + 7 = 1
   \]
2. **At \( x = 0 \) (i.e., \( y_1 \)):**
   \[
   y_1 = f(0) = -6(0)^2 + 7 = 7
   \]
3. **At \( x = 1 \) (i.e., \( y_2 \)):**
   \[
   y_2 = f(1) = -6(1)^2 + 7 = -6 + 7 = 1
   \]

### Step 2: Apply the area formula
Now substitute \( h = 1 \), \( y_0 = 1 \), \( y_1 = 7 \), and \( y_2 = 1 \) into the area formula:

\[
\text{Area} = \frac{1}{3} \left( 1 + 4(7) + 1 \right)
\]
\[
\text{Area} = \frac{1}{3} \left( 1 + 28 + 1 \right)
\]
\[
\text{Area} = \frac{1}{3} \times 30 = 10
\]

### Final Answer:
The enclosed area is **10 square units**.

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Would you like further clarification or more details on any step?

Here are 5 related questions to extend your understanding:
1. How would the area formula change if the limits were \( x = -2 \) and \( x = 2 \)?
2. How does the value of \( a \) in the equation \( y = -ax^2 + bx + c \) affect the shape of the parabola?
3. What happens to the enclosed area if the parabola opens upward instead of downward?
4. Can you generalize the formula for the area under a parabola using integration?
5. How would the area change if \( b \) or \( c \) in the equation \( y = -6x^2 + bx + c \) were non-zero?

**Tip:** In problems involving parabolas, the symmetry around the y-axis often simplifies calculations by reducing the number of points you need to evaluate!

The figure shows the graph of y equals negative ax squared plus bx plus c. Suppose that the points (negative h, y0), (0, y1), and (h, y2) are on the graph. The area enclosed by the parabola, the x-axis, and the lines x equals negative h and x equals h may be given by the formula below. Find the area enclosed by f(x) equals negative 6x squared plus 7, the x-axis, and the lines x equals negative 1 and x equals 1.

Learn how to calculate the area enclosed by the parabola f(x) = -6x^2 + 7, the x-axis, and the lines x = -1 and x = 1. This solution involves using the parabolic area formula and applying key quadratic function concepts. Suitable for Grades 10-12 students.

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