Respuesta :

When we are given a quadratic expression in the standard form:

[tex]y=ax^2+bx+c[/tex]

We can complete the squares to rewrite it as:

[tex]y=a(x+d)^2+e[/tex]

Where:

[tex]\begin{gathered} d=\frac{b}{2a} \\ . \\ e=e-\frac{b^2}{4a} \end{gathered}[/tex]

The poroblem gives us the quadratic equation:

[tex]y=x^2+6x+11[/tex]

Here, we have:

a = 1

b = 6

c = 11

Let's calculate d and e:

[tex]\begin{gathered} d=\frac{6}{2\cdot1}=3 \\ \end{gathered}[/tex][tex]e=11-\frac{6^2}{4\cdot1}=11-\frac{36}{4}=2[/tex]

Then, we can rewrite:

[tex]y=1(x+3)^2+2[/tex]

Thus, the answer is:

[tex]y=(x+3)^2+2[/tex]