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 SolutionWe’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 2Next, 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 3Next, 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 4Finally, all we need to do is multiply the 6 back through the parenthesis to get,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{f\left( x \right) = 6{{\left( {x + 1} \right)}^2} - 3}}\]