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Converting Standard Form to Vertex Form

Standard form can be converted to vertex form by completing the square. This is one of the more complicated conversations, but is nevertheless straightforward. First of all we need a quadratic expression in the standard form (eq. 1).

Eq. 1: \(y=x^2+7x+12\)

With our quadratic expression, we need to put the first two terms into a bracket (eq. 2). These are the basis for the terms that we are going to become inside the bracket in the vertex form.

Eq. 2: \(y=(x^2+7x)+12\)

At this point, you need to divide the contents of the bracket by their greatest common factor to give the a value of the quadratic expression (eq. 3). In this case the value is 1, which means the a coefficient is 1, and is not displayed by convention (though we do so here for clarity), but this is not necessarily the case if the greatest common factor of the contents of the bracket is a different value, and consequently the a coefficient isn't 1.

Eq. 3: \(y=1(x^2+7x)+12\)

Next, we need to find a value to put inside the bracket to make the contents of the bracket a perfect square trinomial. This value is the b value of the standard form, divided by 2, and squared. You then need to multiply this number by the a value you’ve just calculated (which in this case is 1, so no such multiplication needs to take place), and subtract it from the outside of the bracket (eq. 4). You can then simplify the expressions (eq. 5) and like terms can be combined (eq. 6).

Eq. 4: \(y=1(x^2+7x+({7\over2})^2)+12-1({7\over2})^2\)

Eq. 5: \(y=1(x^2+7x+12.25)+12-12.25\)

Eq. 6: \(y=1(x^2+7x+12.25)-0.25\)

The terms in the bracket are now a perfect square trinomial. Following the process to factor a perfect square trinomial, we need a value which is the square root of the c value and half of the b value. This value is then added to the inside of the bracket, along with x to give a perfect square (eq. 6) thus deriving the vertex form of the quadratic expression (eq. 7).

Eq. 7: \(y=(x+3.5)^2-0.25\)

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