Answer:
The required form is \(y-\frac{3}{8}=-\frac{3}{2}(x+\frac{1}{2})^2\)
Step-by-step explanation:
Given : Function \(3x^2+3x+2y=0\)
To find : Complete the square and reduce to one standard form \(y-b=A(x-a)^2 \text{ or } x-a=A(y-b)^2\)?
Solution :
Converting the functio into given standard form,
Taking y one side, \(-2y=3x^2+3x\)
Divide by 3 both side,
\(-\frac{2}{3}y=x^2+x\)
Applying completing the square i.e. add half square both side,
\(-\frac{2}{3}y+(\frac{1}{2})^2=x^2+x+(\frac{1}{2})^2\)
\(-\frac{2}{3}y+(\frac{1}{2})^2=(x+\frac{1}{2})^2\)
\(\frac{2}{3}y-\frac{1}{4}=-(x+\frac{1}{2})^2\)
Divide by \(\frac{2}{3}\) both side,
\(y-\frac{3}{8}=-\frac{3}{2}(x+\frac{1}{2})^2\)
This form relates to \(y-b=A(x-a)^2\)
Where, \(b=\frac{3}{8}, A=-\frac{3}{2} ,a=-\frac{1}{2}\)
Therefore, The required form is \(y-\frac{3}{8}=-\frac{3}{2}(x+\frac{1}{2})^2\)
