What are the coordinates of the vertex for f(x) = x2 + 6x + 13?

If you have
y=ax²+bx+c or
f(x)=ax²+bx+c, then the x value of the vertex is -b/(2a)

so
given that the equation is
f(x)=1x²+6x+13
-b/2a=-6/(2*1)=-6/2=-3

the y value is found by subsituting -3 for x
f(-3)=(-3)²+6(-3)+13
f(-3)=9-18+13
f(-3)=4

the coordainates are (-3,4)

Do you have answer choices?