Question

2x^2 +6x +11 =0In vertex form

Answer 1

The equation was given in its standard form, as follows:

`y=2x^2+6x+11`

To rewrite it in the vertex form, we need to complete the square to write it in the following form:

`y=a(x-h)^2+k`

If we rewrite the standard form using the, we get:

`\begin{gathered} y=2(x^2+3x+(11)/(2))=2(x^2+2* x*(3)/(2)+((3)/(2)_{})^2-((3)/(2))^2+(11)/(2)) \\ y=2((x^2+2* x*(3)/(2)+(9)/(3))-(9)/(4)+(11)/(2))=2((x+(3)/(2))^2-(9)/(4)+(22)/(4))_{} \\ y=2((x+(3)/(2))^2+(13)/(4))=2(x+(3)/(2))^2+(13)/(2) \\ \\ y=2(x+(3)/(2))^2+(13)/(2) \end{gathered}`