Find the vertex of f(x)=4x^2+19x+12. Please help, thanks!

(-19/8, -9/16)

Step-by-step explanation:

Recall that the vertex of a parabola is situated on the axis of symmetry, which here is

      -b         -19

x = = = -2 3/8

      2a           8

To find the y-coordinate of the vertex, evaluate f(x)=4x^2+19x+12 at -19/8.

Using a calculator, we find that f(-19/8) = -10.5625.

Then the vertex is at (-19/8, -10.5625), or (-19/8, -9/16)

idk

step-by-step explanation: i just don't