Question

Use the zeros and the labeled point to write the quadratic functionrepresented by the graph.O A. y=-3x2 - 9x - 6O B. y=-x2 + x + 2O C. y = -2x + 6x - 4O D.y=-3x2 + 3x + 6

Answer 1

`y=-3x^(2)+3x+6`

1) Let's set three equations with these points and solve a system

`\begin{gathered} y=ax^2+bx+c,(2,0) \\ y=ax^2+bx+c,(-1,0) \\ y=ax^2+bx+c,(1,6) \\ \begin{bmatrix}4a+2b+c=0\\ a-b+c=0\\ a+b+c=6\end{bmatrix} \end{gathered}`

Note that the parabola opens down st the coefficient a is negative

`\begin{gathered} 4a+2b+c=0 \\ 4a+2b+c-\left(2b+c\right)=0-\left(2b+c\right) \\ 4a=-\left(2b+c\right) \\ a=-(2b+c)/(4) \\ Plug\:into\:those\:2\:equations\:a=-(2b+c)/(4) \\ (2b+c)/(4)-b+c=0 \\ (-6b+3c)/(4)=0 \\ -6b+3c=0 \\ (2b+c)/(4)+b+c=6 \\ (2b+3c)/(4)=6 \\ 2b+3c=24 \\ -6b+3c=0*(-1) \\ ----- \\ 2b+3c=24 \\ 6b-3c=0 \\ ------ \\ 8b=24 \\ b=3 \\ 6b-3c=0 \\ 6(3)-3c=0 \\ 18=3c \\ c=6 \\ Plug\:them\:into\:that\:and\:solve\:for\:a \\ \begin{equation*} a=-(2b+c)/(4) \end{equation*} \\ a=-(2(3)+6)/(4)=-3 \end{gathered}`

So we can write out the rule:

`y=-3x^2+3x+6`