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How Do You Convert a Quadratic from Standard Form to Vertex Form by Completing the Square?

Write y = x2 + 12x + 32 in vertex form by completing the square.

Summary

  1. Our equation is in standard form to begin with: y=ax2+bx+c
  2. We want to put it into vertex form: y=a(x-h)2+k
  3. We can convert to vertex form by completing the square on the right hand side
  4. 36 is the value for 'c' that we found to make the right hand side a perfect square trinomial
  5. Our perfect square trinomial factors into two identical binomials, (x+6)•(x+6)
  6. The vertex of an equation in vertex form is (h,k), which for our equation is (-6,-4)

Notes

    1. Right now our quadratic equation, y=x2+12x+32 is in standard form
    2. We want to get it into vertex form
    3. To do this, we are going to use the method of completing the square
    1. Standard form of a quadratic equation is y=ax2+bx+c, where 'a' is not 0
    2. Vertex form of a quadratic equation is y=a(x-h)2+k, where (h,k) is the vertex of the quadratic function
    1. 'a', 'b', and 'c' can be any real number, except 'a' cannot be 0
    2. For our equation, a=1, b=12, and c=32
    3. 'h' and 'k' can also be any real number
    4. Vertex form allows us to easily pick out the vertex of a quadratic function
    5. Remember, the vertex is the point at the top or bottom of the graph of a parabola
    1. A perfect square trinomial is a trinomial that can be factored into something squared
    2. So when you factor it, you would have two identical binomials being multiplied together
    1. We want to be able to factor the right hand side into two identical binomials
    2. We can't do that if our last term is 32, so we need to move it over to the other side
    3. Then we can figure out what we DO need to add to get it to factor how we want it
    1. Make sure to subtract 32 on BOTH sides to preserve the equality!
    2. The 32's on the right cancel out, leaving us with just x2+12x
    1. Usually when we are completing the square, we are trying to solve for 'x'
    2. Here we're not actually trying to find a value for 'x' - we're just using the same method to rearrange the equation
    3. Since vertex form also has an 'x' and a 'y', we need to keep that 'y' in our equation while we complete the square on the other side
    1. A perfect square trinomial has the form ax2+bx+c, just like standard form
    2. So 'c' is the constant term we're adding at the end
    3. We want to find a value for 'c' that will allow us to factor the right hand side into two identical binomials
    1. A perfect square trinomial factors into the product of two identical binomials
    2. Since our x2 term doesn't have a coefficient in front of it, we know the first terms of our binomials will be 'x'
    3. Since the middle term of the trinomial is positive, the signs of our binomials will be positive as well
    4. When we multiply out (x+6) times (x+6), we will get 6x+6x for our middle terms, which adds up to 12x
    5. For our last term we would get 6•6, or 36, which is our value for 'c'
    1. If we add 36 to the right hand side, we will have a perfect square trinomial on that side
    2. But if we add it to the right, we need to make sure to add it to the left too!
    3. That way we'll keep both sides of the equation equal
    4. When we simplify the left hand side by adding -32 and 36, we get y+4
    1. Factoring the right hand side will be easy, because we used the factored form to find it in the first place!
    1. We used the factored form of the trinomial to find the value for 'c' in the first place
    2. So we already know what x2+12x+36 factors into!
    3. Since it is a perfect square trinomial, it factors into two identical binomials
    4. Multiplying something by itself is the same thing as squaring it, so we can rewrite (x+6)•(x+6) as (x+6)2
    1. In vertex form, 'y' needs to be by itself on one side
    2. So to get 'y' by itself, we need to subtract 4 from both sides to cancel it out on the left
    1. Make sure to subtract 4 from BOTH sides to preserve the equality
    2. The 4's cancel on the left, leaving 'y' by itself, which is what we wanted!
    3. Remember, vertex form is y=a(x-h)2+k
    4. So here we have a = 1, h = -6, and k = -4
    1. Vertex form is y=a(x-h)2+k
    2. The vertex of an equation in vertex form is (h,k)
    3. In vertex form, we have x MINUS h
    4. So if we're trying to just pick out 'h', we need to flip the sign that's in front of it
    5. So since we have x+6, 'h' is -6
    6. In vertex form we have a PLUS 'k' at the end, which means that it will keep the sign in front of it when we pick it out
    7. So since we're subtracting 4 at the end of our equation, 'k' is -4
    8. That means the vertex is (-6,-4)