Question 281385
There are a couple ways you can go about solving this problem. 

The roots is basically finding when y = 0.

So your equation {{{y = x^2 + 3*x}}} can be written as {{{0 = x^2 + 3*x}}}
Also remember factoring

{{{x^2 + 3x = 0}}}
{{{x(x+3) = 0}}}
{{{x = 0}}} or {{{x+3 = 0}}}

Both roots are:

{{{x = 0 and x = -3}}}

Finding the vertex, min or max is nearly the same process so I'll find two additional points. To find two additional points, pick out any two x values you want. Lets say x = -2 and x = -1. We need to find the y values so:

{{{y = x^2 + 3*x}}}
{{{y = (-2)^2 + 3 * (-2) = 4 - 6 = -2}}} one point is (-2,-2)
{{{y = (-1)^2 + 3 * (-1) = 1 - 3 = -2}}} another point is (-1,-2)

Finally the vertex and min or max. Since this is a parabola which is a U shape graph, you are either going to have a minimum or a maximum that is not infinite. We can tell your parabola opens upward because the coefficient near the {{{x^2}}} is a positive number. It so happens at the vertex will be either a maximum or a minimum and in our case, it will be a minimum (because it will be the lowest point). We can use the vertex formula:

If (h,k) is your vertex, then:

{{{h = -b/2a}}}

where you have a quadratic equation {{{y = ax^2 + bx + c}}}
In our case, {{{h = -3/2}}}, we can find our k by plugging in -3/2 into our equation:

{{{y = (-3/2)^2 + 3(-3/2) = 9/4 - 18/4 = -9/4}}}

Which implies (-3/2,-9/4) is the vertex and with our last statement is a minimum.

Hopes this helps!