Paul's Online Notes
Paul's Online Notes
Home / Algebra / Common Graphs / Parabolas
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 4-2 : Parabolas

9. Convert the following equations into the form \(y = a{\left( {x - h} \right)^2} + k\).

\[f\left( x \right) = 6{x^2} + 12x + 3\]

Show All Steps Hide All Steps

Start Solution

We’ll need to do the modified completing the square process described in the notes for this section.

The first step in this process is to make sure that we have a coefficient of one on the \({x^2}\). So, for this problem that means we need to factor a 6 out of the quadratic to get,

\[f\left( x \right) = 6\left( {{x^2} + 2x + \frac{1}{2}} \right)\]

Be careful to not just cancel out a 6 from each term! We need to factor it out.

Show Step 2

Next, we need to take one-half the coefficient of the \(x\), square it and then add and subtract it onto the equation.

\[{\left( {\frac{2}{2}} \right)^2} = {\left( 1 \right)^2} = 1\] \[\require{color} f\left( x \right) = 6\left( {{x^2} + 2x \,{\color{Red} + 1 - 1} + \frac{1}{2}} \right)\]

Make sure to do the adding/subtracting inside the parenthesis. If we did it outside of the parenthesis we would not be able to do the next step!

Show Step 3

Next, we need to factor the first three terms and combine the last two numbers to get,

\[f\left( x \right) = 6\left( {{{\left( {x + 1} \right)}^2} - \frac{1}{2}} \right)\] Show Step 4

Finally, all we need to do is multiply the 6 back through the parenthesis to get,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{f\left( x \right) = 6{{\left( {x + 1} \right)}^2} - 3}}\]