Quadratic Function

Write a rule for a quadratic function with a graph that has x-intercepts (2,0) and (-6,0) and a maximum point of (-2,4)

quadratic function: $y =ax^2+bx+c$

1. Point (x=2, y=0):

$\quad 0 = a\cdot (2^2)+b\cdot 2 + c \\ \Rightarrow 4a+2b+c=0$

Maximum Point ($x_{max} = -2, \ y_{max} = 4$):

$\begin{array}{|rcll|} \hline x_{max} &=& -\frac{b}{2a} \\ -2 &=& -\frac{b}{2a} \\ 2 &=& \frac{b}{2a} \\ 4a &=& b \\ \mathbf{ a } & \mathbf{=}& \mathbf{\frac{b}{4} }\\ \hline \end{array}$

$\begin{array}{|rcll|} \hline 4a+2b+c&=&0 \quad | \quad a=\frac{b}{4} \\ 4\cdot \frac{b}{4} +2b+c&=&0 \\ b+2b+c&=&0 \\ 3b+c&=&0 \\ 3b&=&-c \\ \mathbf{ b } & \mathbf{=}& \mathbf{-\frac{c}{3}} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline a &=& \frac{b}{4} \quad | \quad b = -\frac{c}{3} \\ a &=& \frac{ -\frac{c}{3}}{4} \\ a &=& -\frac{c}{12} \\\\ y_{max} &=& a\cdot ( x_{max} ) ^2 + b\cdot x_{max} + c \quad | \quad a = -\frac{c}{12} \quad b = -\frac{c}{3} \\ y_{max} &=& -\frac{c}{12}\cdot ( x_{max} ) ^2 -\frac{c}{3}\cdot x_{max} + c \quad | \quad x_{max} = -2\\ y_{max} &=& -\frac{c}{12}\cdot ( -2 ) ^2 -\frac{c}{3}\cdot (-2) + c \quad | \quad y_{max} = 4\\ 4 &=& -\frac{1}{3}\cdot c +\frac{2}{3}\cdot c + c \\ 4 &=& \frac{4}{3}\cdot c \\ \mathbf{ c } & \mathbf{=}& \mathbf{3} \\\\ b&=&-\frac{c}{3}\\ \mathbf{ b } & \mathbf{=}& \mathbf{-1} \\\\ a&=&-\frac{b}{4}\\ \mathbf{ a } & \mathbf{=}& \mathbf{-\frac{1}{4}} \\ \hline \end{array}$

quadratic function:

$\begin{array}{|rcll|} \hline y &=& ax^2+bx+c \\ \mathbf{ y }&\mathbf{=}& \mathbf{-\frac{1}{4}x^2-x+3} \\ \hline \end{array}$