Vertex V = minimum point
is found by x = -b/2a
locate a,b in equation
a= 1 , b= 4
then x = -4/2•1 = -4/2= -2
Now find vertex y = (4ac - b^2)/ 4a^2
. y = ( 4•1•5 - 16)/ 4•1
. y= (20 - 16) / 4
. y = 4/4 = 1
Then Vertex is located at (-2,1)
Now find intercepts x,y
For y- intercept , x= 0, then y = 0+0 + 5 = 5
For x-intercept ,y = 0, then x² + 4x + 5 = 0
BUT , this equation have no x- intercept, because never touches y= 0
