A(x)=-6+x^2+3x

It looks like you have a quadratic function defined by \( A(x) = -6 + x^2 + 3x \). Let's simplify and analyze this function.

1. **Rearranging the Function**:
\[
A(x) = x^2 + 3x - 6
\]

2. **Finding the Vertex**:
The vertex of a parabola given by \( Ax^2 + Bx + C \) can be found using the formula \( x = -\frac{B}{2A} \):
\[
A = 1, \quad B = 3
\]
\[
x = -\frac{3}{2 \cdot 1} = -\frac{3}{2}
\]

To find the y-coordinate of the vertex, substitute \( x = -\frac{3}{2} \) into the function:
\[
A\left(-\frac{3}{2}\right) = \left(-\frac{3}{2}\right)^2 + 3\left(-\frac{3}{2}\right) - 6
\]
\[
= \frac{9}{4} - \frac{9}{2} - 6
\]
\[
= \frac{9}{4} - \frac{18}{4} - \frac{24}{4}
\]
\[
= \frac{9 - 18 - 24}{4} = \frac{-33}{4}
\]

Thus, the vertex is at \( \left(-\frac{3}{2}, -\frac{33}{4}\right) \).

3. **Finding the Roots (if any)**:
To find the roots of \( A(x) = 0 \), we can use the quadratic formula:
\[
x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}
\]
Here, \( A = 1, B = 3, C = -6 \):
\[
x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-6)}}{2 \cdot 1}
\]
\[
= \frac{-3 \pm \sqrt{9 + 24}}{2}
\]
\[
= \frac{-3 \pm \sqrt{33}}{2}
\]

So the roots are:
\[
x = \frac{-3 + \sqrt{33}}{2} \quad \text{and} \quad x = \frac{-3 - \sqrt{33}}{2}
\]

4. **Graphing**:
The graph of the function \( A(x) = x^2 + 3x - 6 \) is a parabola that opens upwards (since the coefficient of \( x^2 \) is positive). The vertex is the lowest point on the graph, and there are two roots indicating that it crosses the x-axis at two points.

If you have any specific question regarding this function or need further analysis, please let me know!