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