The given equation is:

$y = x^2 + 9x - 3$

This is a quadratic equation in the form:

$y = ax^2 + bx + c$

Where:

• $a = 1$ (coefficient of $x^2$),
• $b = 9$ (coefficient of $x$),
• $c = -3$ (constant term).

Key features of the quadratic equation:

1. Vertex: The vertex of the parabola is the point where it changes direction. It can be found using the formula: $x_{\text{vertex}} = -\frac{b}{2a}$ Substituting the values of $a$ and $b$: $x_{\text{vertex}} = -\frac{9}{2 \times 1} = -\frac{9}{2} = -4.5$ To find the $y$-coordinate of the vertex, substitute $x = -4.5$ back into the original equation.

2. Axis of symmetry: The axis of symmetry is the vertical line that passes through the vertex. It is given by $x = -\frac{b}{2a}$, which in this case is $x = -4.5$.

3. Y-intercept: The y-intercept is the point where the graph crosses the y-axis. To find this, set $x = 0$ in the equation: $y = 0^2 + 9(0) - 3 = -3$ So the y-intercept is $(0, -3)$.

4. Direction of the parabola: Since $a = 1 > 0$, the parabola opens upwards.

5. Roots: The solutions to $y = 0$ (where the graph crosses the x-axis) can be found using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Substituting $a = 1$, $b = 9$, and $c = -3$: $x = \frac{-9 \pm \sqrt{9^2 - 4(1)(-3)}}{2(1)} = \frac{-9 \pm \sqrt{81 + 12}}{2} = \frac{-9 \pm \sqrt{93}}{2}$ The roots are: $x = \frac{-9 \pm \sqrt{93}}{2}$

Would you like a more detailed explanation or help with specific points?

Follow-up questions:

1. Would you like to solve for the exact vertex and its corresponding y-coordinate?
2. Should we simplify the roots further or calculate their approximate values?
3. Do you need help graphing this quadratic function?
4. Would you like a breakdown of how to use the quadratic formula?
5. Are you interested in how to factor this equation, if possible?

Tip: The discriminant $b^2 - 4ac$ tells us the nature of the roots. If it's positive, the quadratic has two real roots.