The equation you provided is:

\[
y = -x^2 + 10x - 33
\]

This represents a quadratic equation in the form \( y = ax^2 + bx + c \), where:

- \( a = -1 \)
- \( b = 10 \)
- \( c = -33 \)

### Key Features:

1. **Direction of Parabola**: Since the coefficient of \(x^2\) (i.e., \(a\)) is negative, the parabola opens **downward**.
   
2. **Vertex**: The vertex of a parabola in standard form \( y = ax^2 + bx + c \) can be found using the formula for the x-coordinate of the vertex: 
   \[
   x = -\frac{b}{2a}
   \]
   Plugging in the values:
   \[
   x = -\frac{10}{2(-1)} = 5
   \]
   Now, substitute \(x = 5\) into the original equation to find the y-coordinate:
   \[
   y = -(5)^2 + 10(5) - 33 = -25 + 50 - 33 = -8
   \]
   So, the vertex is at \( (5, -8) \).

3. **Axis of Symmetry**: The axis of symmetry is the vertical line that passes through the vertex. In this case, it is \( x = 5 \).

4. **Y-intercept**: The y-intercept occurs where \(x = 0\):
   \[
   y = -(0)^2 + 10(0) - 33 = -33
   \]
   So, the y-intercept is \( (0, -33) \).

5. **X-intercepts (Roots)**: To find the x-intercepts, we solve \( y = 0 \):
   \[
   -x^2 + 10x - 33 = 0
   \]
   We can use the quadratic formula:
   \[
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   Substituting \(a = -1\), \(b = 10\), and \(c = -33\):
   \[
   x = \frac{-10 \pm \sqrt{10^2 - 4(-1)(-33)}}{2(-1)} = \frac{-10 \pm \sqrt{100 - 132}}{-2}
   \]
   \[
   x = \frac{-10 \pm \sqrt{-32}}{-2}
   \]
   Since the discriminant is negative, the quadratic has no real solutions, meaning the parabola does not cross the x-axis (no real x-intercepts).

Would you like any further details or explanations?

### Related Questions:
1. How do you determine whether a parabola opens up or down?
2. What is the significance of the vertex in a quadratic function?
3. How can you use the quadratic formula to solve for x-intercepts?
4. What does the discriminant tell you about the nature of the solutions?
5. How would the graph change if the coefficient of \(x^2\) were positive?

**Tip**: When the discriminant is negative, the quadratic has no real solutions, meaning the graph does not intersect the x-axis.

Explore how to solve the quadratic equation y = -x^2 + 10x - 33. This problem includes key algebra concepts, such as determining the vertex, finding the axis of symmetry, and solving for the x- and y-intercepts. Suitable for students in Grades 9-10.

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