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The given equation is:

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

Here are some important aspects of this quadratic function:

1. **Vertex Form and Vertex**: To find the vertex of the parabola, you can complete the square or use the formula for the vertex \((h, k)\) where \( h = -\frac{b}{2a} \).
   
   For this equation, \( a = 1 \), \( b = -2 \), and \( c = -10 \).

   \[
   h = -\frac{-2}{2 \times 1} = 1
   \]

   Substitute \( x = 1 \) into the equation to find \( k \):

   \[
   k = (1)^2 - 2(1) - 10 = 1 - 2 - 10 = -11
   \]

   So, the vertex is at \((1, -11)\).

2. **Axis of Symmetry**: The axis of symmetry is the vertical line that passes through the vertex, given by \( x = 1 \).

3. **Y-Intercept**: Set \( x = 0 \) to find the y-intercept:

   \[
   y = 0^2 - 2(0) - 10 = -10
   \]

   The y-intercept is \((0, -10)\).

4. **X-Intercepts**: To find the x-intercepts, set \( y = 0 \):

   \[
   0 = x^2 - 2x - 10
   \]

   Solve this using the quadratic formula:

   \[
   x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-10)}}{2(1)} = \frac{2 \pm \sqrt{4 + 40}}{2} = \frac{2 \pm \sqrt{44}}{2} = \frac{2 \pm 2\sqrt{11}}{2} = 1 \pm \sqrt{11}
   \]

   So, the x-intercepts are \( x = 1 + \sqrt{11} \) and \( x = 1 - \sqrt{11} \).

Would you like more details or have any questions?

### Related Questions:
1. How do you convert a quadratic function into vertex form?
2. What are the effects of changing coefficients \(a\), \(b\), and \(c\) in a quadratic equation?
3. How do you find the focus and directrix of a parabola?
4. What is the significance of the discriminant in a quadratic equation?
5. How can you graph a quadratic function using key points?

**Tip:** Always start by identifying the coefficients when solving a quadratic equation; it helps in applying the right formulas efficiently!

Explore how to solve the quadratic equation y = x^2 - 2x - 10. This example covers key algebraic concepts like finding the vertex, x- and y-intercepts, and using the quadratic formula to find solutions. Suitable for students in Grades 8-10.

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