Solving Quadratic Equations
Converting Standard Form to Vertex Form
Completing the Square
You can covert any quadratic formula from standard form:
y = a Â· x^{ 2} + b
Â· x + c,
to vertex form:
y = a Â· (x  h)^{ 2} + k,
through an algebraic process called completing the square. The next example
demonstrates the steps that are involved in this process.
Example
Convert the quadratic function:
y = 4 Â· x^{ 2} + 6
Â· x + 7,
from standard to vertex form and locate the x and ycoordinates of the vertex.
Solution
Once the formula for the quadratic function has been converted to vertex form:
y = a Â· (x  h)^{ 2} + k,
we can find the vertex by checking the vertex form to find the values of h (which will be
the xcoordinate of the vertex) and k (which will be the ycoordinate of the vertex).
Conversion of the formula from standard to vertex form is a fourstep process called
completing the square.
1. Factor out the coefficient of x2 from all terms.
2. Add and subtract just the right amount1 to create a perfect square
^{1 }To find just the right amount, you take the number that is left multiplying the x after Step 1 has been
completed. Whatever this number is, divide the number by 2 and then take the square of what you are left
with. This is just the right amount to create a perfect square.
3. Factor the perfect square and combine the constants
4. Distribute the factor that is out in front of the equation
The vertex form of the quadratic function is:
This is not quite the same as the â€œclassicâ€ format of the vertex form:
y = a Â· (x  h)^{ 2} + k,
because the number inside the parentheses is added to x, rather than subtracted from x.
To fix this we can use the fact that the negative of a negative is a positive, so that
. Using this to rewrite the vertex form in the â€œclassicâ€ format gives:
With the vertex form in the â€œclassicâ€ format you can go ahead and determine the x and
ycoordinates of the vertex. The xcoordinate is
and the ycoordinate is
