SOLUTION: Write the equation if the quadratic function with roots -9 and a vertex at -6, -1

Since -9 is a root, it has an x-intercept of (-9,0) 

We plot those given points:



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

 
that goes through the vertex.  So we draw that:



And we can sketch in the parabola so that the green line
bisects it:

 
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
 = a

So the equation is

y = (x + 6)² - 1

You can leave it like that or you can continue:

y = (x + 6)(x + 6) - 1

y = (x² + 12x + 36) - 1

y = (x² + 12x + 36) - 

y =  - 

y = 

y = 

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

y = 

Or you can take this:

y = 

write it as

y =  + x + 

and then as

y = x² + x + 3

Edwin