Let f(x)=12x+111ex\displaystyle {f{{\left({x}\right)}}}={12}{x}+{1}-{11}{e}^{{x}}. Then the equation of the tangent line to the graph of f(x)\displaystyle {f{{\left({x}\right)}}} at the point (0,10)\displaystyle {\left({0},-{10}\right)} is given by y=mx+b\displaystyle {y}={m}{x}+{b} for

m=\displaystyle {m}=  
b=\displaystyle {b}=