Question 203715
f(x)= x^2+4x+3

to find the vertex, use  (-b/2a)

-4/2(1) = -4/2 = -2 This is the x part of the vertex (-2,?), then we use -2 in place of x in f(x)

f(-2) = (-2)^2+4(-2)+3
      = 4-8+3
      = -1

Therefore the vertex is (-2,-1)

To find the x - intercepts, you set the equation to equal 0

x^2+4x+3=0 (does it factor?)

(x+3)=0, (x+1)=0
 x= -3,   x=-1

To find the y-intercepts, let x=0

x^2+4x+3
(0)^2+4(0)+3

y=3