График показательной функции \(f\left( x \right) = {a^{x + b}}\) проходит через точки \(\left( { — 1;1} \right)\) и \(\left( { — 3;2} \right)\).
Следовательно: \(\left\{ {\begin{array}{*{20}{c}}{1 = {a^{ — 1 + b}}}\\{2 = {a^{ — 3 + b}}}\end{array}} \right.\)
Разделим первое уравнение на второе: \(\dfrac{{{a^{\, — 1 + b}}}}{{{a^{\, — 3 + b}}}} = \dfrac{1}{2}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,{a^2} = \dfrac{1}{2}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,a = \dfrac{1}{{\sqrt 2 }}.\)
Тогда: \({\left( {\dfrac{1}{{\sqrt 2 }}} \right)^{\, — 1 + b}} = 1\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,{\left( {\dfrac{1}{{\sqrt 2 }}} \right)^{\, — 1 + b}} = {\left( {\dfrac{1}{{\sqrt 2 }}} \right)^0}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\, — 1 + b = 0\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,b = 1.\)
Таким образом: \(f\left( x \right) = {\left( {\dfrac{1}{{\sqrt 2 }}} \right)^{\,x + 1}}\) и \(f\left( { — 9} \right) = {\left( {\dfrac{1}{{\sqrt 2 }}} \right)^{\, — 9 + 1}} = {\left( {\dfrac{1}{{\sqrt 2 }}} \right)^{\, — 8}} = {2^4} = 16.\)
Ответ: 16.
Задача 27. На рисунке изображён график функции \(f\left( x \right) = {a^{x + b}}.\) Найдите \(f\left( { — 9} \right).\)