Question 117704
{{{y = x^2 + 6x + 8}}}
First, it helps if you can picture it. Does it look like a pointy
cup that holds water, or is it a pointy cup that poured all its 
water out?
If the number in front of {{{x^2}}} is positive, it's a pointy cup
that holds water. In this case, The number in front of {{{x^2}}}
is {{{1}}}, since {{{1*x^2 = x^2}}}.
Now you can picture it.
(1) If it touches the x-axis it has 1 real solution for x.
That means you can somehow get {{{x^2 + 6x + 8 = 0}}} into the form 
{{{(x - r)^2 = 0}}}. Then {{{x = r}}} is the only solution.
The other possibilities are:
(2) It floats above the x-axis without touching it.
(3) It crosses the x-axis in 2 places
That's almost all you need to know. Now picture it crossing the x-axis
in 2 places. Where is the vertex? It's exactly in the middle between
the 2 crossing points. The crossing points are the roots
So, if the roots were at {{{x = +1}}} and {{{x = +3}}}, then the vertex
would be at {{{x = +2}}}.
Now all you need to know are the roots of {{{y = x^2 + 6x + 8}}}
They occur at {{{y = 0}}}, so find {{{x}}} when {{{x^2 + 6x + 8 = 0}}}
Right away I see that {{{2 * 4 = 8}}} and {{{2 + 4 = 6}}}.
That leads me to {{{(x + 4)(x + 2) = 0}}}. What makes this true?
Either {{{x = -4}}} or {{{x = -2}}} These are the roots. What's
exactly in the middle? {{{x = -3}}} is right between them.
So, that's where the vertex is. A point needs an x and a y, so what
is the y? Plug {{{x = -3}}} into the original equation
{{{y = x^2 + 6x + 8}}}
{{{y = (-3)^2 + 6*(-3) + 8}}}
{{{y = 9 - 18 + 8}}}
{{{y = -1}}}
So, the vertex is at (-3,-1)
Here's the graph
{{{ graph( 600, 600, -10, 10, -10, 10, x^2 + 6x + 8) }}}