What is the vertex for the equation f(x)=3(x+4)^2-5

Take derivative..
f'(x) = 3*2*(x+4)
f'(x) is must be equal to 0.
6*(x+4)=0
x=-4

for x=-4; what is the y value?
for x=-4; f(x)=-5

the vertex is (-4,-5)

F(x) = 3(x + 4)² - 5
f(x) = 3(x + 4)(x + 4) - 5
f(x) = 3(x² + 4x + 4x + 16) - 5
f(x) = 3(x² + 8x + 16) - 5
f(x) = 3(x²) + 3(8x) + 3(16) - 5
f(x) = 3x² + 24x + 48 - 5
f(x) = 3x² + 24x + 43