f(x)=2(x+2)(x+1) how do i go about doing this problem to find vertex-x and vertex-y

y = 2(x + 1) (x +2) ......let's expand this to get this form ... y = ax^2 + bx + c

y = 2x^2 + 6x + 4  

The x coordinate of the vertex is given by  -b/2a  =  -6/(2*2)  = -6/4 =  -3/2  = -1.5

To find the y coordinate of the vertex, just plug this value into the function and we get

2(-1.5)^2 + 6(-1.5) + 4 = 

2(2.25) - 9  + 4 =

4.5 - 9 + 4 = -0.5

So the vertex is (-1.5, -0,5)

 Here's the graph:    https://www.desmos.com/calculator/sez3bt8lc6

f(x)=2(x+2)(x+1) 

this is a concave up parabola

roots are -2 and -1

axis of symm is    x=(-2-1)/2 = -3/2    this is the x value of the vertex

sub in x=-3/2 to find y value

f(-3/2)=2(1/2)(-1/2)=-1/2

Vertex (-3/2, -1/2)