The fastest way to find the vertex of an quadratic equation in the general from of:

f(x) = ax^{2} + bx +c

is to use the following formula:

x = -b/2a

In this case:

x = -(10)/(2)(1) = -5

Plugging this onto the equation yields:

f(-5) = (-5)^{2} + 10(-5) - 3

f(-5) = 25 - 50 - 3

f(-5) = -28

**The vertex is: (-5, -28)**

The vertex in this case is at the lowest point of the graph and the -28 represents the minimum value of the function.

Mary L.

Thanks!11d