determine in general form the rule of a quadratic function whose zeros are -8 and 2 and whose minimum is -50

Solution: $$ f(x)=y $$ The zeros of a function are also called the roots or the factors. $$ x=-8 $$ $$ x=2 $$ Therefore; $$ y=(x+8)(x-2)+C $$ where C is the constant to meet the minimum value. $$ y=x^2+6x-16+C $$ Make the equation in the form: $$ y=a(x-h)^2+k $$ Where k is the minimum value of the function. Therefore; Make it a perfect square trinomial by completing the square... $$ y+16+(\frac{6}{2})^2=x^2+6x+(\frac{6}{2})^2+C $$ $$ y+16+9=x^2+6x+9+C $$ $$ y+25=(x+3)^2+C $$ $$ y=(x+3)^2+(C-25) $$ $$ k=(C-25) $$ $$ -50=C-25 $$ $$ -50+25=C $$ $$ C=-25 $$ Going back to our original equation... $$ y=x^2+6x-16+C $$ $$ f(x)=x^2+6x-16-25 $$ $$ f(x)=x^2+6x-41 $$ ANSWER: $$ f(x)=x^2+6x-41 $$