We’ll use a sample equation.......12X^2 + 8X - 10 = Y

1. Start with the pattern

of the Vertex Form.................................A(x + h)^2 + K = Y

2. From the original equation,

think of the X-variable

terms as a separate entity......(12X^2 + 8X) - 10 = Y

3. Factor out the leading

coefficient, so that the

coefficient on the X^2

is just 1..............................12(X^2 + 8/12 X) - 10 = Y

I’d reduce that improper

fraction...............................12(X^2 + 2/3 X) - 10 = Y

4. Fill in “A” in the

Vertex Form..........................................12(X + h)^2 + K = Y

5. Fill in the “h” value

of the Vertex Form

with 1/2 the new

coefficient on X in

Step 3..............................................12(X + 2/6)^2 + K = Y

I like smaller numbers, so I’m reducing

that pesky improper fraction................12(X + 1/3)^2 + K = Y

6. Think for a second about what we’ve

done. By creating the (X + h) binomial

and squaring it, we’ve actually added

more to the equation. (FOIL it through

if you need proof.) So we need to

remove it again.

Square “h”, multiply it by “A”,

and subtract it from the

constant in the General

Form equation...................12(X + 1/3)^2 + (- 10 - 12/9) = Y

I feel I’m doing more reducing that any

real work here.....................12(X + 1/3)^2 + (- 10 - 4/3) = Y

7. A little arithmetic and,

TAH-DAH, I’m ready

to solve for X and label

the vertex on a graph......................12(X + 1/3)^2 - 34/3 = Y

*Many texts add the step of finding the perfect square trinomial before factoring it into the binomial squared. I think it’s just a way to insure that you subtract out the extraneous constant value but I prefer "the elegant solution."*

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