Factored form can be converted directly into to vertex form. This is using the principle that the factored form describes the x intercepts of the parabola, while the vertex form describes the vertex of the parabola and the shape of it. Taking an equation in the factored form as an example (Eq. 1), we can follow this process.

Eq. 1: \(y=(x+4)(x+3)\)

The a, h and k values of the vertex form relate to the vertex, so we can use these to build the equation. The first thing we can calculate is the h value of the equation, as this is going to be x coordinate of the vertex, and equidistant from both of the x intercepts. This is calculated by adding the x intercepts together and dividing by 2 (Eq. 2). This gives the h value of 3.5.

Eq. 2: \(x={4+3\over2}=3.5\)

This makes the h value of the vertex form 3.5. Next, we need to find the y value of the vertex, which will give the k value in the vertex form. To do this, we put the h value into the factored form equation to find the y value of the parabola, which we can do by putting the h value of the vertex form into the place of the x value of the factored form of the equation (Eq. 3). This gives the k value of -0.25.

Eq. 3: \(y=(3.5+4)(3.5+3)=0.5\times-0.5=-0.25\)

This has given us the h and k values of the vertex form of the equation of 3.5 and -0.25 respectively. The last value required is the a value. The a value is a measure of how much the parabola differs from a x² trend. To do this, we need to find a second point on the parabola along with the vertex. To do this, we use one of the x intercepts. We need to subtract the y value of the vertex from 0 (the y value of the x intercept) to get the actual change in the parabola (Eq. 4). Next, we need to find the expected change in the parabola were the tend is x², this is by squaring the change in the x axis (0.5) (Eq. 5).

Eq. 4: \(\Delta x_O=0--0.25=0.25\)

Eq. 5: \(\Delta x_E=0.5^2=0.25\)

To derive a point on an expected parabola, we first need to find the difference between the x value of the vertex and that of one of the x-intercepts (eq. 3). If you square this value, you have the expected change in y value given that distance from the vertex along the x axis along the parabola (eq. 4).

Eq. 6: \(-3.5-3=-0.5\)

Eq. 7: \(-0.5^2=0.25\)

Next, you need the observed change in y on the parabola at the distance along the x axis that you’ve measured. As these points are the x-intercepts and the vertex, you can simply subtract the y coordinate of the vertex (-0.25) from the y coordinate of the x-intercept (which will always be 0) (eq. 5).

Eq. 8: \(0--0.25=-0.25\)

Finally, when you have your observed and expected parabolas, you can divide the observed value by your expected value to give you the degree of compression in the parabola (eq. 6).

Eq. 9: \({0.25\over0.25}=1\)

This is calculation for the a value (eq. 7), which means the a, h and k values have been determined. These can be substituted into the general vertex form to give the complete equation in the vertex form (eq. 8).

Eq. 10: \(y=(x+3.5)^2-0.25\)

This thus completes determining the quadratic equation in the vertex form.