График функции \(f\left( x \right) = a\,tgx + b\) проходит через точки \(\left( {0; — \frac{3}{2}} \right)\) и \(\left( {\frac{\pi }{4}; — 2} \right)\). Следовательно:
\(\left\{ {\begin{array}{*{20}{c}}{ — \frac{3}{2} = a\,tg0 + b}\\{ — 2 = a\,tg\frac{\pi }{4} + b}\end{array}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{ — \frac{3}{2} = a \cdot 0 + b}\\{ — 2 = a \cdot 1 + b}\end{array}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{b = — \frac{3}{2}\,\,\,\,\,\,\,\,\,\,}\\{a + b = — 2}\end{array}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,a = — \frac{1}{2}.} \right.} \right.} \right.\)
Ответ: – 0,5.
Задача 20. На рисунке изображён график функции \(f\left( x \right) = a\,\,{\text{tg}}\,x + b.\) Найдите a.