What is the vertex form \[y = {x^2} + 8x + 15\]?

Hint: Recollect the vertex formula. Express the given equation into an algebraic identity.
The vertex form is given by \[y = {(x - h)^2} + k\] , where $h$,$k$ are the vertex.

Complete step by step answer:
Let the vertex be \[\left( {h,{\text{ }}k} \right)\] .
The vertex form is given by \[y = {(x - h)^2} + k\] , where $ h$,$k$ are the vertex.
The given equation is\[y = {x^2} + 8x + 15\]
Express the quadratic equation in square notation
That is, in the form of \[{(a + b)^2}\]
\[{(a + b)^2} = {a^2} + 2ab + {b^2}\]
 Comparing with the expansion and given equation ,
\[{a^2} = {x^2}\]
\[a = x\]
\[2ab = 8x = 2(x)(4)\]
Here $b$ turns to be $4$;
By comparing the middle term, we get the value of $b$.
Hence add and subtract $16$ so as not to change the equation meaning
\[{x^2} + 8x + 15\] =\[{x^2} + 2(4)(x) + {4^2} - 16 + 15\]
Compare the equation with the standard expansion
\[{x^2} + 8x + 15\] = \[{(x + 4)^2} - 1\]
(After combining the square term left part is \[ - 16{\text{ }} + 15{\text{ }} = {\text{ }} - 1\])
\[y = {(x + 4)^2} - 1\]
Comparing the above equation with the vertex form \[y = {(x - h)^2} + k\]
We can write \[y = {(x - ( - 4))^2} + ( - 1)\]
Hence we can write \[h{\text{ }} = {\text{ }} - 4\] and \[k{\text{ }} = {\text{ }} - 1\]
The vertex form is \[y = {(x + 4)^2} - 1\]

Additional information: The vertex form of a parabola's equation is generally expressed as: $y = a{(x - h)^2} + k$ . \[\left( {h,{\text{ }}k} \right)\]is the vertex. If$a$ is positive then the parabola opens upwards like a regular "U". If$a$is negative, then the graph opens downwards like an upside down "U".
You can draw the parabola and show vertex to attract the examiner.

Note: As all terms are positive in the given equation we are expressing as \[{\left( {a{\text{ }} + {\text{ }}b} \right)^2}\] notation. If the middle term is negative then express as\[{\left( {a{\text{ }} + {\text{ }}b} \right)^2}\].
When you are adding a term to the equation you must subtract it also, that keeps the equation unchanged.