Question 287399
The best way to think of it is:
The axis of symmetry is {{{x = k}}} where {{{k}}}
is midway between the roots of the equation.
Then you find the vertex by plugging in {{{x=a}}}
back into the equation to get {{{y}}} , then {{{x,y}}}
is the vertex.
An easy formula is: if the equation is in the form
{{{y = ax^2 + bx + c}}}, then the axis of symmetry is at
{{{x = -b/(2a)}}}
The y-intercept is at {{{y = c}}}
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I'll do your 1st problem:
{{{y = x^2 + 6x - 11}}} 
The y-intercept is at {{{y = -11}}}
Now find roots:
{{{x^2 + 6x  - 11 = 0}}}
{{{x^2 + 6x = 11}}}
{{{x^2 + 6x + (6/2)^2 = 11 + (6/2)^2}}}
{{{x^2 + 6x + 9 = 11 + 9}}}
{{{(x + 3)^2 = 20}}}
{{{x + 3 = sqrt(20)}}}
{{{x = -3 +-sqrt(20)}}}
The axis of symmetry is midway between the 2 roots which are:
root1 ={{{ -3 + sqrt(20)}}}
root2 = {{{-3 - sqrt(20)}}}
{{{-3}}} is midway between these roots, so
{{{x = -3}}} is the axis of symmetry
And the simple formula {{{x = -b/(2a)}}} gives the same answer:
{{{a = 1}}}
{{{b = 6}}}
{{{x = -6/(2*1)}}}
{{{x = -3}}}
Now plug this result back into the equation:
{{{y = x^2 + 6x - 11}}}
{{{y = (-3)^2 + 6*(-3) - 11}}}
{{{y = 9 - 18 - 11}}}
{{{y = -20}}}
So, the vertex is at (-3,-20)
I'll plot the equation:
{{{ graph( 500, 500, -10, 10, -20, 20, x^2 + 6x - 11) }}}