Question 83166
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1) write an equation for the parabola whose 
vertex is at (-8,4) and passes through (-6,-2) 

The equation of a parabola with vertex (h,k) is

    y = a(x - h)² + k

Substitute in (h,k) = (-8,4)

    y = a(x - (-8) )² + (4)

    y = a(x + 8)² + 4

Now we can substitute (x,y) = (-6,-2)

 (-2) = a(-6 + 8)² + 4

   -2 = a(2)² + 4

   -2 = a(4) + 4

   -2 = 4a + 4

   -6 = 4a

 {{{-6/4}}} = a 

 {{{-3/2}}} = a

So substitute {{{-3/2}}} for a in

    y = a(x + 8)² + 4

and we have

    y = {{{-3/2}}}(x + 8)² + 4

Drawing the graph:

{{{drawing(400,300,-11,1,-4,5,

   graph(400,300,-11,1,-4,5,-1.5(x+8)^2+4),
  locate (-8.1,4.25,o), locate(-6.1,-1.8,o), locate(-5.7,-1.7,"(-6,-2)"),
  locate(-8.7,4.5,"(-8,4)") 

 )}}}

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2) i need to write: y = x² + 4x - 1 in vertex form 

    y = x² + 4x - 1

factor 1 out of the first two terms on the right:

    y = 1(x² + 4x) - 1

Complete the square by 
1. taking 1/2 of the coefficient of x
2. squaring that quantity
3. Adding it and subtracting it inside the
   parentheses.

1. 1/2 of 4 is 2
2. 2 squared is 4
3.

    y = 1(x² + 4x + 4 - 4) - 1

Change the parentheses to brackets:

    y = 1[x² + 4x + 4 - 4] - 1

Factor the frirst three terms inside the
bracket:

    y = 1[(x + 2)(x + 2) - 4] - 1

Write (x + 2)(x + 2) as (x + 2)²

    y = 1[(x + 2)² - 4] - 1

Remove the bracket by distributing the 1
into the bracket, leaving the parentheses
intact.

    y = 1(x + 2)² - 4 - 1

Combine the last two terms

    y = 1(x + 2)² - 5

Copare that to

    y = a(x - h)² + k

and you can see that a = 1, h = -2, k = -5

so the vertex is (h,k) = (-2,-5) 

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3) which quadratic function has its vertex at (-2,7)
 and opens down?

Well, many many poarabolas have that vertex and
open down.  We'll find one.

  Substitute (h,k) = (-2,7) in the standard equation

     y = a(x - h)² + k

     y = a(x - (-2) )² + 7
     
     y = a(x + 2)² + 7

But then choose any negative number for a, and its
graph will open downward.  For instance, letting 
a = -1 gives this parabola:

    y = -(x + 2)² + 7

{{{drawing(246.2,400,-6,2,-4,9, locate(-2.14,7.3,o),

  locate(-3,8,"(-2,7)"),
   graph(246.2,400,-6,2,-4,9,-(x+2)^2+7) 

 )}}} 

Edwin</pre>