جواب کار در کلاس صفحه 67 درس 2 ریاضی و آمار دهم انسانی (تابع)

الف

\(\begin{array}{l}y = {x^2} \Rightarrow \left\{ \begin{array}{l}a = 1\\\\b = 0\\\\c = 0\end{array} \right.\\\\ \Rightarrow {x_s} = \frac{{ - b}}{{2a}} = \frac{{ - 0}}{{2 \times 1}} = 0\\\\ \Rightarrow {y_s} = {x_s}^2 = {(0)^2} = 0\end{array}\)

\(:\,S = (0\,,\,0)\) رأس

ب

\(\begin{array}{l}y = 4 - {x^2} \Rightarrow \left\{ \begin{array}{l}a = - 1\\\\b = 0\\\\c = 4\end{array} \right.\\\\ \Rightarrow {x_s} = \frac{{ - b}}{{2a}} = \frac{{ - 0}}{{2 \times ( - 1)}} = 0\\\\ \Rightarrow {y_s} = 4 - {x_s}^2 = 4 - {(0)^2} = 4\end{array}\)

\(:\,S = (0\,,\,4)\) رأس

پ

\(\begin{array}{l}y = 2{x^2} - 4x + 1 \Rightarrow \left\{ \begin{array}{l}a = 2\\\\b = - 4\\\\c = 1\end{array} \right.\\\\ \Rightarrow {x_s} = \frac{{ - b}}{{2a}} = \frac{{ - ( - 4)}}{{2 \times (2)}} = \frac{4}{4} = 1\\\\ \Rightarrow {y_s} = 2{x_s}^2 - 4{x_s} + 1 = \\\\2{(1)^2} - 4(1) + 1 = 2 - 4 + 1 = - 1\end{array}\)

\(:\,S = (1\,,\,1)\) رأس

ت

\(\begin{array}{l}y = 2{(x - 1)^2} + 1 = 2({x^2} - 2x + 1) + 1 = \\\\2{x^2} - 4x + 2 + 1 = 2{x^2} - 4x + 3\\\\ \Rightarrow \left\{ \begin{array}{l}a = 2\\\\b = - 4\\\\c = 3\end{array} \right.\\\\ \Rightarrow {x_s} = \frac{{ - b}}{{2a}} = \frac{{ - ( - 4)}}{{2 \times (2)}} = \frac{4}{4} = 1\\\\ \Rightarrow {y_s} = 2{x_s}^2 - 4{x_s} + 3 = \\\\2{(1)^2} - 4(1) + 3 = 2 - 4 + 3 = 1\end{array}\)

\(:\,S = (1\,,\,1)\) رأس