Question

Достаточно сложная , нужно полное решение с объяснением. за эту максимальное кол-во .

Answer 1

$y=\frac{11}{8}|x| - \frac{3}{8}|x-8|-x-1$

Объяснение:

$y=a|x|+b|x-8|+kx+c$

точки

(-2;0)

(0;-4)

(8;2)

(9;2)

$\begin{cases}2a + 10b - 2k + c = 0\\8b + c = -4\\8a + 8k + c = 2\\9a + b + 9k + c = 2\end{cases}$

$\begin{cases}2a + 10b - 2k + c = 0\\c = -8b-4\\8a + 8k + c = 2\\9a + b + 9k + c = 2\end{cases}$

$\begin{cases}2a + 10b - 2k -8b-4 = 0\\c = -8b-4\\8a + 8k -8b-4 = 2\\9a + b + 9k -8b-4 = 2\end{cases}$

$\begin{cases}2a + 2b - 2k - 4 = 0\\c = -8b-4\\8a + 8k -8b = 2+4\\9a - 7b + 9k - 4 = 2\end{cases}$

$\begin{cases}2a + 2b - 2k - 4 = 0\ /:2\\c = -8b-4\\8a + 8k -8b = 6\ /:2\\9a - 7b + 9k = 2+4\end{cases}$

$\begin{cases}a + b - k - 2 = 0\\c = -8b-4\\4a + 4k -4b = 3\\9a - 7b + 9k = 6\end{cases}$

$\begin{cases}a =-b+k+2\\c = -8b-4\\4a + 4k -4b = 3\\9a - 7b + 9k = 6\end{cases}$

$\begin{cases}a =-b+k+2\\c = -8b-4\\4(-b+k+2) + 4k -4b = 3\\9(-b+k+2) - 7b + 9k = 6\end{cases}$

$\begin{cases}a =-b+k+2\\c = -8b-4\\-4b+4k+8 + 4k -4b = 3\\-9b+9k+18 - 7b + 9k = 6\end{cases}$

$\begin{cases}a =-b+k+2\\c = -8b-4\\- 8b + 8k + 8 = 3\\- 16b + 18k + 18 = 6\ /:2\end{cases}$

$\begin{cases}a =-b+k+2\\c = -8b-4\\- 8b + 8k + 8 = 3\\-8b+9k +9 = 3\end{cases}$

$\begin{cases}a =-b+k+2\\c = -8b-4\\- 8b = 3-8k-8\\-8b+9k = 3-9\end{cases}$

$\begin{cases}a =-b+k+2\\c = -8b-4\\- 8b = -8k-5\\-8b+9k =-6\end{cases}$

$\begin{cases}a =-b+k+2\\c = -8b-4\\- 8b = -8k-5\\-8k-5+9k = -6\end{cases}$

$\begin{cases}a =-b+k+2\\c = -8b-4\\- 8b = -8k-5\\k = -6+5\end{cases}$

$\begin{cases}a =-b+k+2\\c = -8b-4\\- 8b = -8k-5\\k = -1\end{cases}$

$\begin{cases}a =-b-1+2\\c = -8b-4\\- 8b = -8 \cdot (-1)-5\\k = -1\end{cases}$

$\begin{cases}a =-b+1\\c = -8b-4\\- 8b = 8-5\\k = -1\end{cases}$

$\begin{cases}a =-b+1\\c = -8b-4\\- 8b = 3\ /:(-8)\\k = -1\end{cases}$

$\begin{cases}a =-b+1\\c = -8b-4\\b = - \frac{3}{8} \\k = -1\end{cases}$

$\begin{cases}a =-\left(- \frac{3}{8} \right) +1\\c = -8 \cdot \left( - \frac{3}{8} \right) -4\\b = - \frac{3}{8} \\k = -1\end{cases}$

$\begin{cases}a = \frac{3}{8} +1\\c =3 -4\\b = - \frac{3}{8} \\k = -1\end{cases}$

$\begin{cases}a = \frac{11}{8}\\c = -1\\b = - \frac{3}{8} \\k = -1\end{cases}$