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Let's solve the given equation \( y = -2x^2 + 16x - 27 \) and plot the five points on the parabola as requested, including the vertex and two points on each side.

### Step 1: Find the vertex
The equation is in the standard quadratic form \( y = ax^2 + bx + c \). Here, \( a = -2 \), \( b = 16 \), and \( c = -27 \).

The x-coordinate of the vertex is given by the formula:
\[
x = \frac{-b}{2a} = \frac{-16}{2(-2)} = \frac{-16}{-4} = 4
\]

Substitute \( x = 4 \) back into the equation to find the y-coordinate:
\[
y = -2(4)^2 + 16(4) - 27 = -2(16) + 64 - 27 = -32 + 64 - 27 = 5
\]

So, the vertex is at \( (4, 5) \).

### Step 2: Find two points to the left and right of the vertex
Let’s choose values for \( x \) around the vertex (say, \( x = 2, 3 \) to the left and \( x = 5, 6 \) to the right).

#### For \( x = 3 \):
\[
y = -2(3)^2 + 16(3) - 27 = -2(9) + 48 - 27 = -18 + 48 - 27 = 3
\]
So, \( (3, 3) \) is a point.

#### For \( x = 2 \):
\[
y = -2(2)^2 + 16(2) - 27 = -2(4) + 32 - 27 = -8 + 32 - 27 = -3
\]
So, \( (2, -3) \) is a point.

#### For \( x = 5 \):
\[
y = -2(5)^2 + 16(5) - 27 = -2(25) + 80 - 27 = -50 + 80 - 27 = 3
\]
So, \( (5, 3) \) is a point.

#### For \( x = 6 \):
\[
y = -2(6)^2 + 16(6) - 27 = -2(36) + 96 - 27 = -72 + 96 - 27 = -3
\]
So, \( (6, -3) \) is a point.

### Step 3: Summary of Points
- Vertex: \( (4, 5) \)
- Points: \( (3, 3), (2, -3), (5, 3), (6, -3) \)

You can now plot these points on the graph provided.

Would you like further details or have any other questions?

### Related Questions:
1. How can we determine if a parabola opens upwards or downwards?
2. What is the axis of symmetry for this parabola?
3. How do you find the y-intercept of a quadratic function?
4. How do the coefficients \( a \), \( b \), and \( c \) affect the shape and position of the parabola?
5. Can we solve for the x-intercepts (roots) of this parabola?

**Tip**: The sign of the coefficient \( a \) determines whether the parabola opens upwards (\( a > 0 \)) or downwards (\( a < 0 \)).

Plot five points on the parabola of the equation y = -2x^2 + 16x - 27, including the vertex and two points to the left and right.

Learn how to plot five points on the parabola of the quadratic equation y = -2x^2 + 16x - 27. This solution includes finding the vertex and calculating points on both sides of the parabola for a complete graph. Suitable for high school algebra students in Grades 9-10.

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