SOLUTION: 1) write an equation for the parabola whose vertex is at (-8,4) and passes through (-6,-2) 2) i need to write: y=x square + 4x - 1 in vertex form 3) which quadratic function

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: 1) write an equation for the parabola whose vertex is at (-8,4) and passes through (-6,-2) 2) i need to write: y=x square + 4x - 1 in vertex form 3) which quadratic function       Log On


   

Question 83166: 1) write an equation for the parabola whose vertex is at (-8,4) and passes through (-6,-2)
2) i need to write: y=x square + 4x - 1 in vertex form
3) which quadratic function has its vertex at (-2,7) and opens down?

thank you
Found 2 solutions by Mona27, Edwin McCravy:
Answer by Mona27(45) About Me  (Show Source):
You can put this solution on YOUR website!
The general expression for a quadratic in completed square form is:
y=a%28x%2Bb%29%5E2%2Bc
where the vertex would be (-b,c).
1) Since the vertex is (-8,4), then the formula should be:
y=a%28x%2B8%29%5E2%2B4
And to find the value of a, substitute the other point (-6,-2) in the equation:
-2=4a%2B4
giving the value of a as -3%2F2=-1.5
2) To change a normal quadratic into the completed square form, first take half the coefficient of x (2) and place it in a bracket like this:
%28x%2B2%29%5E2
Now this expression gives you x%5E2%2B4x%2B4 and since we only need the first two terms, we need to eliminate the last one. To do that you simply subtract 4:
%28x%2B2%29%5E2-4
And finally place the last term right after that.
%28x%2B2%29%5E2-4-1=%28x%2B2%29%5E2-5 This means the vertex is (-2,-5)
3) Same as the first one, and to make a quadratic function "open down" you will need to put a negative sign before the bracket:
-%28x%2B2%29%5E2%2B7
or
7-%28x%2B2%29%5E2

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!

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%2F4 = a 

 -3%2F2 = a

So substitute -3%2F2 for a in

    y = a(x + 8)² + 4

and we have

    y = -3%2F2(x + 8)² + 4

Drawing the graph:



----------------------------------------------

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) 

---------------------------------------------------

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

 

Edwin