How do you write $y = - 10{x^2} + 2$ in vertex form?

Hint: We have to rewrite the given equation using some algebraic steps in order to make it look like the Standard Vertex form of the quadratic function.
For this the first step is to isolate the ${x^2}$ terms by moving $2$ to the other side of the equal sign.
Then, we need a leading coefficient of $1$ for completing the square so factor out the current leading coefficient of $ - 10$. Next, we have to create a perfect square.
Isolate the $y$-term. So move $2$ to the other side of the equal sign. The equation obtained will be the vertex form of the given equation.

Formula used: Vertex form of the quadratic function:
$y = a{\left( {x - h} \right)^2} + k$ where $\left( {h,k} \right)$ is the vertex or the “center” of the quadratic function or the parabola.

Complete step-by-step solution:
Given quadratic function: $y = - 10{x^2} + 2$
This quadratic equation is in the form$y = a{x^2} + bx + c$.
However, we need to rewrite it using some algebraic steps in order to make it look like this
$y = a{\left( {x - h} \right)^2} + k$
This is the vertex form of the quadratic function where $\left( {h,k} \right)$ is the vertex or the “center” of the quadratic function or the parabola.
Here, the coefficient of${x^2}$,$a = - 10$ not equal to$1$.
Since we will be completing the square, we will isolate the ${x^2}$ terms.
So, the first step is to move $2$ to the other side of the equal sign.
Subtract $2$ from both sides of the equation $y = - 10{x^2} + 2$.
$ \Rightarrow y - 2 = - 10{x^2}$
Now, we need a leading coefficient of $1$ for completing the square so factor out the current leading coefficient of $ - 10$.
$ \Rightarrow y - 2 = - 10\left( {{x^2}} \right)$
Now, we have to create a perfect square.
$ \Rightarrow y - 2 = - 10{\left( {x - 0} \right)^2}$
Isolate the $y$-term. So move $2$ to the other side of the equal sign.
Add $2$ to both sides of the equation.
$ \Rightarrow y = - 10{\left( {x - 0} \right)^2} + 2$

Therefore, the given equation in vertex form is $y = - 10{\left( {x - 0} \right)^2} + 2$.

Note: We can also solve this type of question using Sneaky Tidbit Method.
When working with the vertex form of a quadratic function: $h = \dfrac{{ - b}}{{2a}}$ and $k = f\left( h \right)$
The “$a$” and “$b$” referenced here refers to $f\left( x \right) = a{x^2} + bx + c$.
Compare $y = - 10{x^2} + 2$ with $f\left( x \right) = a{x^2} + bx + c$ and determine the value of $a$, $b$ and $c$.
Here, $y = - 10{x^2} + 2$ can be written as $y = - 10{x^2} + 0x + 2$.
So, $a = - 10$, $b = 0$ and $c = 2$.
Now, find the value of $h$ by putting the values of $a$ and $b$ in $h = \dfrac{{ - b}}{{2a}}$.
Here, $a = - 10$ and $b = 0$.
So, $h = \dfrac{{ - 0}}{{2 \times \left( { - 10} \right)}}$
On simplification, we get
$ \Rightarrow h = 0$
Now, find $k$ by putting the value of $h$ in $f\left( x \right)$.
$ \Rightarrow k = f\left( 0 \right)$
Put $f\left( x \right) = a{x^2} + bx + c$ in the above equation.
$ \Rightarrow k = a{\left( 0 \right)^2} + b\left( 0 \right) + c$
Put the values of $a$, $b$ and $c$ in the above equation.
Here, $a = - 10$, $b = 0$ and $c = 2$.
$ \Rightarrow k = \left( { - 10} \right){\left( 0 \right)^2} + 0\left( 0 \right) + 2$
On simplification, we get
$k = 2$
Therefore, the vertex is $\left( {h,k} \right) = \left( {0,2} \right)$.
Now, put the value of$a$, $h$ and $k$ in Standard Vertex form of the quadratic function.
Since, Vertex form of the quadratic function:
$y = a{\left( {x - h} \right)^2} + k$ where $\left( {h,k} \right)$ is the vertex or the “center” of the quadratic function or the parabola.
Here, $a = - 10$, $h = 0$ and $k = 2$.
Therefore, the given equation in vertex form is $y = - 10{\left( {x - 0} \right)^2} + 2$.