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Sprawdź, czy punkt P należy do wykresu funkcji f.
a) [tex]f(x) = \frac{6}{x - 4} , P(2,-4)[/tex]
b) [tex]f(x) = \frac{3}{x+2} , P(4, \frac{1}{2} )[/tex]
c) [tex]f(x) = \frac{3}{x - 1} , P(\frac{1}{2} , -\frac{3}{2} )[/tex]
d)[tex]f(x) = \frac{0,5}{x + 4} , P(-6, -\frac{1}{4} )[/tex]

Odpowiedź :

Basetlaviz

[tex]a) \ f(x) = \frac{6}{x-4}, \ \ P(2, -4)\\\\L = -4\\\\P = \frac{6}{2-4} = \frac{6}{-2} = -3\\\\L\neq P, \ \ NIE[/tex]

[tex]b) \ f(x) = \frac{3}{x+2}, \ \ P(4, \frac{1}{2})\\\\L = \frac{1}{2}\\\\P = \frac{3}{4+2} = \frac{3}{6} = \frac{1}{2}\\\\L = P, \ \ TAK[/tex]

[tex]c) \ f(x) = \frac{3}{x-1}, \ \ P(\frac{1}{2}, -\frac{3}{2})\\\\L = -\frac{3}{2}\\\\P = \frac{3}{\frac{1}{2}-1} =\frac{3}{-\frac{1}{2}} = -6\\\\L \neq P, \ \ NIE[/tex]

[tex]d) \ f(x) = \frac{0,5}{x+4}, \ \ P(-6, -\frac{1}{4})\\\\L = -\frac{1}{4}\\\\P = \frac{0,5}{-6+4} = \frac{0,5}{-2} = -\frac{1}{4}\\\\L = P, \ \ TAK[/tex]

a)   f(x)=6/(x-4) , P=(2,-4)

f(2)=6/(-2-4)=6/(-6)=-1

-1 ≠ -4

Nie.

b)  f(x)=3/(x+2) , P=(4,1/2)

f(4)=3/(4+2)=3/6=1/2

1/2 = 1/2

Tak.

c)   f(x)=3/(x-1) , P=(1/2,-3/2)

f(1/2)=3/(1/2-1)=3/(-1/2)=-6

-6 ≠ -3/2

Nie.

d)    f(x)=0,5/(x+4)  , P=(-6,-1/4)

f(-6)=0,5/(-6+4)=0,5/(-2)=-1/4

-1/4 = -1/4

Tak.

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