Question 661828
<pre>
Since -9 is a root, it has an x-intercept of (-9,0) 

We plot those given points:

{{{drawing(400,250,-13,3,-3,7,
graph(400,250,-13,3,-3,7),
circle(-6,-1,.1),circle(-9,-0,.1),
locate(-8,-1.4,"Vertex(-6,-1)"), locate(-9.2,.8,"(-9,0)"))

)}}}

We know that the axis of symmetry is a vertical line through
the vertex:

 
that goes through the vertex.  So we draw that:

{{{drawing(400,250,-13,3,-3,7,
graph(400,250,-13,3,-3,7),
circle(-6,-1,.1),circle(-9,-0,.1), green(line(-6,-10,-6,10)),
locate(-8,-1.4,"Vertex(-6,-1)"), locate(-9.2,.8,"(-9,0)"))

)}}}

And we can sketch in the parabola so that the green line
bisects it:
{{{drawing(400,250,-13,3,-3,7,
graph(400,250,-13,3,-3,7,(1/9)(x+9)(x+3)),
circle(-6,-1,.1),circle(-9,-0,.1), green(line(-6,-10,-6,10)),
locate(-8,-1.4,"Vertex(-6,-1)"), locate(-9.2,.8,"(-9,0)"))

)}}}
 
All quadratic functions have the equation

y = a(x - h)² + k

where the vertex is (h,k).  In this case (h,k) = (-6,-1), so the

equation is

y = a(x - (-6))² + (-1), or upon simplifying,

y = a(x + 6)² - 1

We only need the value of a.

It goes through (-9,0) so we substitute x=-9 and y=0

y = a(x + 6)² - 1
0 = a(-9 + 6)² - 1
0 = a(-3)² - 1
0 = a(9) - 1
0 = 9a - 1
1 = 9a
{{{1/9}}} = a

So the equation is

y = {{{1/9}}}(x + 6)² - 1

You can leave it like that or you can continue:

y = {{{1/9}}}(x + 6)(x + 6) - 1

y = {{{1/9}}}(x² + 12x + 36) - 1

y = {{{1/9}}}(x² + 12x + 36) - {{{9/9}}}

y = {{{(x^2 + 12x + 36)/9}}} - {{{9/9}}}

y = {{{(x^2 + 12x + 36 - 9)/9}}}

y = {{{(x^2 + 12x + 27)/9}}}

From there you can either factor the numerator
and leave it like this:

y = {{{((x+3)(x+9))/9}}}

Or you can take this:

y = {{{(x^2 + 12x + 27)/9}}}

write it as

y = {{{x^2/9}}} + {{{12/9}}}x + {{{27/9}}}

and then as

y = {{{1/9}}}x² + {{{4/3}}}x + 3

Edwin</pre>