График показательной функции \(f\left( x \right) = {a^x} + b\) проходит через точки \(\left( {0; — 3} \right)\) и \(\left( { — 2; — 1} \right)\). Следовательно:
\(\left\{ {\begin{array}{*{20}{c}}{ — 3 = {a^0} + b}\\{ — 1 = {a^{ — 2}} + b}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{b = — 4\,\,\,\,\,\,}\\{ — 1 = \frac{1}{{{a^2}}} + b}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\,\frac{1}{{{a^2}}} = 3\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,a = \frac{1}{{\sqrt 3 }}.\)
Таким образом: \(f\left( x \right) = {\left( {\frac{1}{{\sqrt 3 }}} \right)^x} — 4\) и \(f\left( { — 8} \right) = {\left( {\frac{1}{{\sqrt 3 }}} \right)^{ — 8}} — 4 = 81 — 4 = 77.\)
Ответ: 77.