Question

What are the vertex and x-intercepts of the graph of the function below? y =(x + 4)(x + 2)

Answer 1

y = x^2 -4x -21 = x^2 -4x +4 -25 = (x-2)^2 - 25. 
Now minimum y = - 25 and occurs at x = 2. 
Therefore vertex is at (2,-25). We may factor y = (x-2-5)(x-2+5) = (x-7)(x+3). x-intercepts occur at those values of x where y =0.
Now y = 0 for x = -3 or 7.

Answer 2

In the form y=ax^2+bx+c
the x value of the vertex is -b/2a

the x intercept is where th e linecrosses the x axis or where y=0


expand
(x+4)(x+2)=x^2+6x+8
a=1
b=6
x value of vertex=-6/2(1)=-6/2=-3
sub back to find y value
(-3)^2+6(-3)+8
9-18+8
-1
y value is -1
vertex is (-3,-1)


x intercept
set to zero
0=(x+4)(x+2)
set each to zero
0=x+4
-4=x
x+2=0
x=-2
x intercept are at (-4,0) and (-2,0)



vertex is (-3,-1)
x ints are at (-4,0) and (-2,0)