Question

How do you write the quadratic equation \[y=3{{x}^{2}}-12x+4\] in vertex form?

Answer 1

Hint: Compare the given quadratic equation with the general form given as: - \[y=a{{x}^{2}}+bx+c\]. Find the respective values of a, b and c. Now, find the discriminant of the given quadratic equation by using the formula: - \[D={{b}^{2}}-4ac\], where ‘D’ is the notation for the discriminant. Now, use the relation: - \[y=a\left[ {{\left( x+\dfrac{b}{2a} \right)}^{2}}-\dfrac{D}{4{{a}^{2}}} \right]\] to write the quadratic equation in vertex form.

Complete step by step answer:
Here, we have been provided with the quadratic equation: - \[y=3{{x}^{2}}-12x+4\] and we are asked to write it in vertex form.
Now, we are going to use the method of completing the square to solve the question. This method states that if we have a quadratic equation of the form \[y=a{{x}^{2}}+bx+c\] then its vertex form is written as: - \[y=a\left[ {{\left( x+\dfrac{b}{2a} \right)}^{2}}-\dfrac{D}{4{{a}^{2}}} \right]\], where \[D={{b}^{2}}-4ac\], called the discriminant of the quadratic equation.
So, on comparing the given quadratic expression: - \[y=3{{x}^{2}}-12x+4\] with the general form given as: - \[y=a{{x}^{2}}+bx+c\], we can conclude that: -
\[\Rightarrow \] a = 3, b = -12 and c = 4
Applying the formula for discriminant of a quadratic equation given as: - \[D={{b}^{2}}-4ac\], where ‘D’ is the discriminant, we get,
\[\begin{align}
  & \Rightarrow D={{\left( -12 \right)}^{2}}-4\times 3\times 4 \\
 & \Rightarrow D=144-48 \\
 & \Rightarrow D=96 \\
\end{align}\]
Now, \[y=a{{x}^{2}}+bx+c\] can be written as: -
\[\begin{align}
  & \Rightarrow y=a\left( {{x}^{2}}+\dfrac{b}{a}x+\dfrac{c}{a} \right) \\
 & \Rightarrow y=a\left( {{x}^{2}}+2.\dfrac{b}{2a}x+{{\left( \dfrac{b}{2a} \right)}^{2}}-\dfrac{{{b}^{2}}}{{{\left( 2a \right)}^{2}}}+\dfrac{c}{a} \right) \\
\end{align}\]
\[\Rightarrow y=a\left[ {{\left( x+\dfrac{b}{2a} \right)}^{2}}-\left( \dfrac{{{b}^{2}}}{{{\left( 2a \right)}^{2}}}-\dfrac{c}{a} \right) \right]\]
\[\Rightarrow y=a\left[ {{\left( x+\dfrac{b}{2a} \right)}^{2}}-\left( \dfrac{{{b}^{2}}-4ac}{4{{a}^{2}}} \right) \right]\]
Substituting, \[{{b}^{2}}-4ac=0\], we get,
\[\Rightarrow y=a\left[ {{\left( x+\dfrac{b}{2a} \right)}^{2}}-\dfrac{D}{4{{a}^{2}}} \right]\]
So, we have derived a general form for the vertex representation of a quadratic expression. Therefore, for the expression: - \[y=3{{x}^{2}}-12x+4\], we have, substituting the values of a, b, c and D.
\[\begin{align}
  & \Rightarrow y=3\left[ {{\left( x+\left( \dfrac{-12}{2\times 3} \right) \right)}^{2}}-\dfrac{96}{4\times {{\left( 3 \right)}^{2}}} \right] \\
 & \Rightarrow y=3\left[ {{\left( x-2 \right)}^{2}}-\dfrac{8}{3} \right] \\
\end{align}\]
Hence, the above expression represents the quadratic expression in vertex form.

Note: You may note that we have derived a general expression for the vertex form. You must remember the result of this vertex form because in coordinate geometry of parabola we will be asked to find the vertex of parabola which can easily be done by this expression. The vertex is given as \[\left( \dfrac{-b}{2a},\dfrac{-D}{4a} \right)\]. You may see that we have applied the method of completing the square for the derivation. This method is generally used for the derivation. This method is generally used for finding the roots of a quadratic equation.