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The given equation is \( y = -2x^2 + 8x - 5 \), which represents a parabola. To find the vertex of this parabola, we use the vertex formula for a quadratic equation \( y = ax^2 + bx + c \). The x-coordinate of the vertex is given by:

\[
x = \frac{-b}{2a}
\]

In this case:
- \( a = -2 \)
- \( b = 8 \)

Substituting these values:

\[
x = \frac{-8}{2(-2)} = \frac{-8}{-4} = 2
\]

Now, substitute \( x = 2 \) into the original equation to find the y-coordinate of the vertex:

\[
y = -2(2)^2 + 8(2) - 5
\]
\[
y = -2(4) + 16 - 5 = -8 + 16 - 5 = 3
\]

Thus, the vertex of the parabola is \( (2, 3) \).

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### Related Questions:
1. What is the axis of symmetry for this parabola?
2. How do you determine if a parabola opens upward or downward?
3. Can we express the given quadratic equation in vertex form?
4. How would you find the x-intercepts of this parabola?
5. What is the maximum or minimum value of the function?

**Tip:** The vertex of a parabola is either its maximum or minimum point, depending on the direction the parabola opens.

Find the vertex of the parabola y = -2x^2 + 8x - 5

Learn how to find the vertex of the parabola y = -2x^2 + 8x - 5 with a step-by-step solution. This problem covers important algebraic concepts, including the vertex formula and quadratic equations. Suitable for students in Grades 9-12.

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