Bestimme die Asymptoten:
f(x)=1xf(x)=\frac1xf(x)=x1
f(x)=1−1xf\left(x\right)=1-\frac1xf(x)=1−x1
f(x)=1+xxf\left(x\right)=\frac{1+x}xf(x)=x1+x
f(x)=x1+xf(x)=\frac x{1+x}f(x)=1+xx
f(x)=x21+x2f(x)=\frac{x^2}{1+x^2}f(x)=1+x2x2
f(x)=∣x∣1+xf(x)=\frac{\left|x\right|}{1+x}f(x)=1+x∣x∣
f(x)=1+1xf(x)=1+\frac1xf(x)=1+x1
f(x)=x+1xf(x)=x+\frac1xf(x)=x+x1
f(x)=x−1xf(x)=x-\frac1xf(x)=x−x1
f(x)=x2+1xf(x)=x^2+\frac1xf(x)=x2+x1
f(x)=x+1x2f(x)=x+\frac1{x^2}f(x)=x+x21
f(x)=3x−3x2−2x+2f\left(x\right)=\dfrac{3x-3}{x^2-2x+2}f(x)=x2−2x+23x−3
f(x)=x2−1x2−2x+1f\left(x\right)=\frac{x^2-1}{x^2-2x+1}f(x)=x2−2x+1x2−1
f(x)=2x3−x2x3+x2f(x)=\frac{2x^3-x^2}{x^3+x^2}f(x)=x3+x22x3−x2
f(x)=x+∣x∣∣x∣−1f(x)=\frac{x+\left|x\right|}{\left|x\right|-1}f(x)=∣x∣−1x+∣x∣
f(x)=−4x3+8x2+23x9x2−18x+9f(x)=\frac{-4x^3+8x^2+23x}{9x^2-18x+9}f(x)=9x2−18x+9−4x3+8x2+23x
f(x)=(x+3)2⋅(x2−1)(x+3)⋅(x2+1)f(x)=\frac{\left(x+3\right)^2\cdot(x^2-1)}{\left(x+3\right)\cdot(x^2+1)}f(x)=(x+3)⋅(x2+1)(x+3)2⋅(x2−1)
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