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The equation provided is \( y = x^2 + 5x - 6 \), and we are tasked with finding the vertex algebraically.

To find the vertex of a quadratic equation in the form \( y = ax^2 + bx + c \), we can use the formula for the x-coordinate of the vertex:

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

In this equation:

- \( a = 1 \) (the coefficient of \( x^2 \)),
- \( b = 5 \) (the coefficient of \( x \)).

Now we can plug these values into the formula:

\[
x_{\text{vertex}} = -\frac{5}{2(1)} = -\frac{5}{2}
\]

Next, to find the y-coordinate of the vertex, substitute \( x = -\frac{5}{2} \) into the original equation:

\[
y = \left(-\frac{5}{2}\right)^2 + 5\left(-\frac{5}{2}\right) - 6
\]

Let me calculate that for you:

\[
y = \frac{25}{4} - \frac{25}{2} - 6
\]

Simplifying:

\[
y = \frac{25}{4} - \frac{50}{4} - \frac{24}{4} = \frac{25 - 50 - 24}{4} = \frac{-49}{4}
\]

Thus, the vertex is:

\[
\left( -\frac{5}{2}, -\frac{49}{4} \right)
\]

The vertex of the quadratic function \( y = x^2 + 5x - 6 \) is \( \left( -\frac{5}{2}, -\frac{49}{4} \right) \).

Would you like more details on the steps? Also, here are some related questions:

1. How does the vertex relate to the graph of a quadratic equation?
2. Can the vertex be used to determine the axis of symmetry?
3. What is the significance of the coefficient \( a \) in determining the direction of the parabola?
4. How can we complete the square to find the vertex?
5. How do we calculate the y-intercept of the quadratic equation?

**Tip:** For any quadratic equation, the vertex helps determine the maximum or minimum point of the parabola, depending on whether it opens upward or downward.

Find the vertex for the function y = x^2 + 5x - 6 algebraically.

Learn how to algebraically find the vertex of the quadratic function y = x^2 + 5x - 6. This problem covers key algebraic concepts such as finding the vertex using the formula x = -b/2a and substituting into the equation to find the y-coordinate.

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