Question

What is the vertex of, $y=2{{\left( x-3 \right)}^{2}}+1$ ?

Answer 1

Hint: First of all we would identify the equation. The equation given to us is the equation of a parabola. Here, we will use the formula for writing the equation of a parabola in its vertex form. Then, we will compare the standard equation with the equation given to us. This will give us the required vertex of the parabola.

Complete step-by-step answer:
The expression given to us in the problem is:
$\Rightarrow y=2{{\left( x-3 \right)}^{2}}+1$
Let us name the above equation as (1), so we have:
$\Rightarrow y=2{{\left( x-3 \right)}^{2}}+1$ ........ (1)

Now, we will make use of the vertex form of writing an equation of parabola. The equation of a standard parabola in the vertex form is written as:
$\Rightarrow y=a{{\left( x-h \right)}^{2}}+k$
Let us name the above equation as (2), so we have:
$\Rightarrow y=a{{\left( x-h \right)}^{2}}+k$ ........ (2)

In the above equation,
‘a’ is the focal length of the parabola. And,
$\left( h,k \right)$ is the vertex of the parabola.

On comparing the standard equation with the equation given to us, we see that our equation is already in the vertex form.
Thus, on comparing equation number (1) and (2), we get:
$\begin{align}
  & \Rightarrow h=3 \\
 & \Rightarrow k=1 \\
\end{align}$

Thus, the coordinate of the vertex of the given parabola equation comes out to be $\left( 3,1 \right)$.
Hence, the vertex of the parabola, $y=2{{\left( x-3 \right)}^{2}}+1$, comes out to be $\left( 3,1 \right)$.

Note: The equation given to us was already in the vertex form. Had it been in the factored form, that is: $y=a\left( bx+c \right)\left( dx+e \right)$, or in the form of a standard quadratic equation, that is, $y=a{{x}^{2}}+bx+c$ , then the vertex could be calculated by equating the ‘y’ term as zero. Then the x-coordinate of the vertex would be $-\dfrac{b}{2a}$. And putting this value in our equation would give the y-coordinate of the vertex.