Question

How do you write $y = 3{x^2} - 12x + 1$ in vertex form?

Answer 1

Hint: We have to write the quadratic equation $y = 3{x^2} - 12x + 1$ in the vertex form. The vertex form of a quadratic function is defined by \[f\;(x){\text{ }} = \;a{(x{\text{ }} - {\text{ }}h)^2}\; + \;k\] , where \[(h,{\text{ }}k)\] is the vertex of the parabola.
To convert the equation into vertex form first subtract $1$ from both the sides of the equation.
Now, make the right hand side of the equation a perfect square in terms of the variable $x$.
The resultant equation is the standard form of $y = 3{x^2} - 12x + 1$ in vertex form.

Complete step-by-step answer:
Consider the given equation is $y = 3{x^2} - 12x + 1$.
We have to convert the equation into vertex form.
The vertex form of a quadratic function is given as;
\[f\;(x){\text{ }} = \;a{(x{\text{ }} - {\text{ }}h)^2}\; + \;k\]
where \[(h,{\text{ }}k)\] is the vertex of the parabola.
Subtract $1$ from both the sides of the equation $y = 3{x^2} - 12x + 1$ we get,
$y - 1 = 3{x^2} - 12x$
Take $3$common from the right side of the equation.
$y - 1 = 3({x^2} - 4x)$
Form a perfect square in the right-hand side of the equation for that add and subtract $4$ in the bracket.
$y - 1 = 3({x^2} - 4x + 4 - 4)$
Here, we can write ${x^2} - 4x + 4$ as ${(x - 2)^2}$.
$y - 1 = 3[{(x - 2)^2} - 4]$
$ \Rightarrow y - 1 = 3{(x - 2)^2} - 12$
Now, add $1$ from both sides of the equation,
$ \Rightarrow y = 3{(x - 2)^2} - 12 + 1$
$ \Rightarrow y = 3{(x - 2)^2} - 11$
Compare the equation with \[f\;(x){\text{ }} = \;a{(x{\text{ }} - {\text{ }}h)^2}\; + \;k\], we have \[(h,{\text{ }}k) = (2, - 11)\] is the vertex of the parabola.

Final Answer: The vertex form $y = 3{x^2} - 12x + 1$ is $y = 3{(x - 2)^2} - 11$.

Note:
The vertex form of a quadratic function is defined by \[f\;(x){\text{ }} = \;a{(x{\text{ }} - {\text{ }}h)^2}\; + \;k\] , where \[(h,{\text{ }}k)\] is the vertex of the parabola.
\[(h,{\text{ }}k)\] is the vertex of the parabola, and $h = k$ is the axis of symmetry.
• Here, $h$ represents a horizontal shift (how far left, or right, the graph has shifted from \[x\; = {\text{ }}0\] ).
• The $k$ represents a vertical shift (how far up, or down, the graph has shifted from \[y\; = {\text{ }}0\]).
• The $h$ value is subtracted in this form, and that the $k$ value is added. If the equation is \[y\; = {\text{ }}2{\left( {x\; - {\text{ }}1} \right)^2}\; + {\text{ 2}}\] , the value of $h$ is $1$ , and $k$ is $2$ .
 If the equation is \[y\; = {\text{ }}3{\left( {x\; + {\text{ }}4} \right)^2}\; - {\text{ }}6\] , the value of $h$ is\[ - 4\] , and $k$ is $ - 6$ .