The given equation is:

\[
y = -x^2 - x + 3
\]

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

- \( a = -1 \) (coefficient of \( x^2 \))
- \( b = -1 \) (coefficient of \( x \))
- \( c = 3 \) (constant)

### Key Features of the Parabola

1. **Direction**: Since the coefficient of \( x^2 \) (i.e., \( a \)) is negative, the parabola opens **downwards**.

2. **Vertex**: The vertex form of a quadratic function can be found using the formula for the \( x \)-coordinate of the vertex:

\[
x_{\text{vertex}} = \frac{-b}{2a} = \frac{-(-1)}{2(-1)} = \frac{1}{-2} = -\frac{1}{2}
\]

Substitute \( x = -\frac{1}{2} \) into the original equation to find the corresponding \( y \)-coordinate:

\[
y_{\text{vertex}} = -\left(-\frac{1}{2}\right)^2 - \left(-\frac{1}{2}\right) + 3
\]

\[
y_{\text{vertex}} = -\frac{1}{4} + \frac{1}{2} + 3 = \frac{-1 + 2 + 12}{4} = \frac{13}{4} = 3.25
\]

Thus, the vertex is \( \left( -\frac{1}{2}, 3.25 \right) \).

3. **Y-intercept**: The y-intercept occurs when \( x = 0 \). Substituting \( x = 0 \) into the equation:

\[
y = -(0)^2 - (0) + 3 = 3
\]

So, the y-intercept is \( (0, 3) \).

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

\[
0 = -x^2 - x + 3
\]

Multiply the entire equation by \( -1 \) to simplify:

\[
x^2 + x - 3 = 0
\]

Now, solve this quadratic equation using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where \( a = 1 \), \( b = 1 \), and \( c = -3 \):

\[
x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-3)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 12}}{2} = \frac{-1 \pm \sqrt{13}}{2}
\]

Thus, the solutions are:

\[
x_1 = \frac{-1 + \sqrt{13}}{2}, \quad x_2 = \frac{-1 - \sqrt{13}}{2}
\]

### Summary:
- The parabola opens downwards.
- Vertex: \( \left( -\frac{1}{2}, 3.25 \right) \)
- Y-intercept: \( (0, 3) \)
- X-intercepts: \( x_1 = \frac{-1 + \sqrt{13}}{2} \), \( x_2 = \frac{-1 - \sqrt{13}}{2} \)

Would you like more details or have any questions?

### Follow-up Questions:
1. How would the graph of the parabola change if \( a \) were positive?
2. What is the significance of the discriminant in the quadratic formula?
3. How would you convert this equation to its vertex form?
4. What is the axis of symmetry for the parabola?
5. How do the x-intercepts relate to the solutions of the quadratic equation?

**Tip**: When solving quadratic equations, always check if they can be factored easily before using the quadratic formula.

Explore the solution to the quadratic equation y = -x² - x + 3. Learn how to find the vertex, the y-intercept, and the x-intercepts, using key algebraic formulas. Suitable for students in Grades 9-10 studying quadratic equations and parabolas.

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