Finding the vertex of praobola

I know theres two ways to find it by using these two equations i just forget how to do it. Examples would be great! Ill show u equations f(x)=a(x-h)^2+k and -b/2a, f(-b/2a) nm figured it out i think. f(x)=x^2+5x+4 (-5/2 thats what i get for the first equation now how do i get the second one i know you replug -5/2 for x but im lost in fractions and i hate fractions!

In vertex form: y= a(x-h)^2 + k, the vertex is (h,k)

In root form: (x-r1)(x-r2), r1 is one root (where in the parabola crosses the x axis) and r2 is another root. The vertex is between them and up or down a certain number. The least you can get, for me, is the where the coordinate is for x, but no y-value. You need a calculator for root form.

The -5/2 is the x value, so -b/2a works well!

To remove fractions, multiply the equation by the common denominator. In this case, it is 2. You'll be rid of fractions until your answer.

A second way is to change it into a decimal. -5/2 is also -2.5.

f(-2.5)= (-2.5)^2 + 5(-2.5) + 4
f(-2.5)= 6.25 - 12.5 +4
f(-2.5)= -6.25 + 4
f(-2.5)= -2.25

You can always change it back to fractions from here.

thx man i owe u