Question

How do you find the vertex of $ y=3{{x}^{2}}+12x-6 $ ?

Answer 1

Hint: To find the vertex of the given equation first we will compare the given equation with the standard quadratic equation which is given as $ y=a{{x}^{2}}+bx+c;a\ne 0 $. Then we will calculate the X-coordinate by using the formula $ \dfrac{-b}{2a} $ then by substituting the value of x into the given equation we get the value of Y-coordinate.

Complete step by step answer:
We have been given an equation $ y=3{{x}^{2}}+12x-6 $
We have to find the vertex of the given equation.
As the given equation $ y=3{{x}^{2}}+12x-6 $ is of the form $ y=a{{x}^{2}}+bx+c;a\ne 0 $
On comparing the equations we get a=3, b=12 and $ c=-6 $
Then we know that vertex is the highest or lowest point of the equation and is given as $ \left( x,y \right).

Now, we know that the X-coordinate of the vertex is given by the quadratic formula $ \dfrac{-b}{2a} $.
Now, substituting the values in the above formula we get
 $ \begin{align}
  & \Rightarrow x=\dfrac{-b}{2a} \\
 & \Rightarrow x=\dfrac{-12}{2\times 3} \\
\end{align} $
Now, simplifying the obtained equation we get
 $ \begin{align}
  & \Rightarrow x=\dfrac{-12}{6} \\
 & \Rightarrow x=-2 \\
\end{align} $
Now, substituting the obtained value of x in the given equation we get
 $ \begin{align}
  & y=3{{x}^{2}}+12x-6 \\
 & \Rightarrow y=3\times {{\left( -2 \right)}^{2}}+12\times \left( -2 \right)-6 \\
\end{align} $
Now, simplifying the above equation further we get
 $ \begin{align}
  & \Rightarrow y=3\times 4+24-6 \\
 & \Rightarrow y=12-24-6 \\
 & \Rightarrow y=12-30 \\
 & \Rightarrow y=-18 \\
\end{align} $
So we get a vertex of $ y=3{{x}^{2}}+12x-6 $ as $ \left( -2,-18 \right) $ .

Note:
The vertex of a quadratic equation is the highest or lowest point of that equation. The point lies on the plane of symmetry of the entire parabola. In order to find the vertex of the quadratic equation, we can use either the quadratic formula or complete the square method. In this question, we use quadratic formula.