Zad. 1.60
Wykresem funkcji liniowej f(x)=ax + b jest prosta nachylona do osi OX pod kątem α, gdzie a = tgα
a)
[tex]f(x) = \sqrt{3}x + 2 \\ a = \sqrt{3} \\ tg\alpha = \sqrt{3} \\ \alpha = 60^o[/tex]
b)
[tex]f(x) = x - \frac{\sqrt{3}}{3} \\ a =1 \\ tg\alpha =1 \\ \alpha = 45^o[/tex]
c)
[tex]f(x) = \frac{- \sqrt{3}}{3}x - 1 \\ a =\frac{- \sqrt{3}}{3} \\ tg\alpha =-\frac{\sqrt{3}}{3} \\ tg\alpha =- tg 30^o \\ tg\alpha =tg (180^o-30^o) \\ tg\alpha =tg150^o \\ \alpha =150^o[/tex]
d)
[tex]f(x) =- \sqrt{3}x - 1 \\ a =- \sqrt{3} \\ tg\alpha =- \sqrt{3} \\ tg\alpha =- tg 60^o \\ tg\alpha =tg (180^o-60^o) \\ tg\alpha =tg120^o \\ \alpha =120^o[/tex]
e)
[tex]f(x) =1+ \frac{\sqrt{3}}{3}x \\ a =\frac{\sqrt{3}}{3} \\ tg\alpha =\frac{\sqrt{3}}{3} \\ \alpha =30^o[/tex]
f)
[tex]f(x) = \sqrt{3} - x \\ a =-1 \\ tg\alpha =-1 \\ tg\alpha =- tg 45^o \\ tg\alpha =tg (180^o-45^o) \\ tg\alpha =tg135^o \\ \alpha =135^o[/tex]
