a)
[tex]A(-2,-1)\\B(3,2)\\y = ax+b\\\left \{ {{-1=-2a+b} \atop {2=3a+b}} \right. \\\left \{ {{b=2a-1} \atop {2=3a+2a-1}} \right. \\5a=3\\a=\frac{3}{5} \\b= 2*\frac{3}{5} -1=\frac{1}{5}\\ y=\frac{3}{5}x +\frac{1}{5}[/tex]
Mz: [tex]x=b=\frac{1}{5}[/tex]
[tex]0=\frac{3}{5} x+\frac{1}{5} \\0=3x+1\\x=-\frac{1}{3}[/tex]
Zatem zbiorem argumentów, dla których funkcja przyjmuje wartości nieujemne jest:
[tex]x\geq -\frac{1}{3}[/tex]
b)
[tex]A(1,2)\\B(4,4)\\y=ax+b\\\left \{ {{4=4a+b} \atop {2=a+b}} \right. \\\left \{ {{4-4a=b} \atop {2=a+4-4a}} \right. \\-3a=-2\\a=\frac{2}{3} \\b=4-4*\frac{2}{3} =\frac{12-8}{3} =\frac{4}{3} \\y=\frac{2}{3}x+\frac{4}{3}[/tex]
Mz: [tex]x=b=\frac{4}{3}[/tex]
[tex]f(x)\geq 0\\\frac{2}{3} x+\frac{4}{3} \geq 0\\2x+4\geq 0\\x\geq -2\\[/tex]
c)
[tex]A(-4,2)\\B(2,-1)\\\left \{ {{2=-4a+b} \atop {-1=2a+b}} \right. \\\left \{ {{2+4a=b} \atop {-1=2a+2+4a}} \right. \\6a=-3\\a=-\frac{1}{2} \\b=0\\y=-\frac{1}{2}x\\[/tex]
Mz: [tex]x=b=0\\[/tex]
[tex]f(x)\geq 0\\-\frac{1}{2} x\geq 0\\x\leq 0[/tex]
d)
[tex]A(0,3)\\B(2,-3)\\\left \{ {{3=b} \atop {-3=2a+b}} \right. \\-3=2a+3\\2a=-6\\a=-3\\y=-3x+3\\[/tex]
Mz: [tex]x=b=3[/tex]
[tex]f(x)\geq 0\\-3x+3\geq 0\\3x\leq 3\\x\leq 1[/tex]
