Given:
The equation is
[tex]y=x^2+2x+3[/tex]
To find:
The vertex form of the given equation, i.e., [tex]y=x^2+2x+3[/tex].
Solution:
We have,
[tex]y=x^2+2x+3[/tex]
It can be written as
[tex]y=(x^2+2x)+3[/tex]
Add and subtract square of half of coefficient of x, i.e., [tex]\left(\dfrac{2}{2}\right)^2=1[/tex].
[tex]y=(x^2+2x+1)-1+3[/tex]
[tex]y=(x^2+2(x)(1)+1^2)+2[/tex]
[tex]y=(x+1)^2+2[/tex] [tex][\because (a+b)^2=a^2+2ab+b^2][/tex]
Therefore, the required form of given function is [tex]y=(x+1)^2+2[/tex].