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f(2) = 0 i f(0) = 4
y = ax + b - równanie kierunkowe prostej
Tworzymy układ równań podstawiając za x 2 i y 0 (z f(2) = 0)
oraz za x 0 i y 4 (z f(0) = 4)
0 = 2a + b
4 = 0*a + b
2a + b = 0
b = 4
2a + 4 = 0
2a = -4 /:2
a = -2
f(x) = -2x + 4
b)
f(⁵/₂) = ¹/₄ i f(2) = -1
y = ax + b
¹/₄ = ⁵/₂ a + b |*4
-1 = 2a + b |*(-4)
10a + 4b = 1
-8a - 4b = 4
-------------------- +
2a = 5 /:2
a = ⁵/₂
b = -1 - 2 * (⁵/₂) = -1 - 5
b = -6
y = ⁵/₂ x - 6
8.
y = ax + b
a = 0 - funkcja stała
a)
f(x) = (m+2)x + 7
a = m + 2
m + 2 = 0
m = -2
b)
f(x) = (2 + 2/m)x + 4
a = 2 + 2/m
m ≠ 0
2 + 2/m = 0 |*m
2m + 2 = 0
2m = -2 /:2
m = -1