Question

How do you write \[f\left( x \right)=2{{x}^{2}}+x-6\] in vertex form?

Answer 1

Hint: In the above type of question that has been mentioned we need to change the format of the equation from the given form to its vertex form for which first we will write down the required form of vertex equation i.e. the general equation then we will be able to see from the equation that we need to find the vertex point (h,k) which can be derived from the formula \[x=-\frac{b}{2a}\] where x is the x co-ordinate of the vertex then we will use this value in the given quadratic equation to find value of k which we will use it in the vertex form of the given equation and the equation formed will be our vertex form of the given quadratic equation.

Complete step by step answer:
In the above question that has been stated where we need to find the vertex from for the given equation for this we are going to first write the general equation of vertex which is:
\[y=a{{\left( x-h \right)}^{2}}+k\]
Now to write any quadratic equation in vertex form we need to find the vertex co-ordinates which are h and k. To find those co-ordinates we will fist find the x co-ordinate of the vertex which can be found with the help of the formula where \[x=-\dfrac{b}{2a}\] in this formula x is the x co-ordinate of the vertex a is the coefficient of the first term of the quadratic equation i.e. \[{{x}^{2}}\] and b is the coefficient of the second term of the quadratic equation i.e. x, when we substitute the coefficient values in the formula we will get the x co-ordinate of the vertex. Now we can find the values of “a'' and “b” from the equation that has been mentioned in the question and we get it as \[a=2\] and \[b=1\] by substituting this we will get the value of x co-ordinate of the vertex which will be
\[\begin{align}
  & \Rightarrow {{x}_{vertex}}=-\dfrac{1}{2\left( 2 \right)} \\
 & \Rightarrow {{x}_{vertex}}=-\dfrac{1}{4} \\
\end{align}\]
Now that we have got the x co-ordinate of the vertex we need to find the y co-ordinates of the vertex which can be found by substituting the value of x co-ordinate of the vertex in the quadratic equation mentioned in the question and we will get the y co-ordinate of the vertex as:
\[\begin{align}
  & \Rightarrow {{y}_{vertex}}=2{{\left( -\dfrac{1}{4} \right)}^{2}}+\left( -\dfrac{1}{4} \right)-6 \\
 & \Rightarrow {{y}_{vertex}}=-\dfrac{49}{8} \\
\end{align}\]
Now that we know both the co-ordinates of the vertex we can get the equation of vertex form.
As we know that h and k are none other than the x and y co-ordinate of vertex point we will substitute it in the general equation of vertex form which will result us with the final vertex equation.
So the final equation of the vertex is:
\[\begin{align}
  & \Rightarrow y=1{{\left( x-\left( -\dfrac{1}{4} \right) \right)}^{2}}+\left( -\dfrac{49}{8} \right) \\
 & \Rightarrow y={{\left( x+\dfrac{1}{4} \right)}^{2}}-\dfrac{49}{8} \\
\end{align}\]
The vertex form of the quadratic equation formed is \[y={{\left( x+\dfrac{1}{4} \right)}^{2}}-\dfrac{49}{8}\].


Note:
This method is not the only method through which you will be able to get the vertex form of equation we clearly have seen that only the vertex co-ordinates are required and quadratic equation can also be converted to parabolic equation and the vertex of that parabola is the same as the required vertex so we just need to convert the quadratic equation into its standard form i.e. \[4a\left( y-b \right)={{\left( x-c \right)}^{2}}\] where c and b are the required x and y co-ordinates of the vertex respectively.