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Find the relative position of the asymptote y=0y=0y=0 to the curve of f(x)=2x3−x2+1x2+1f\left(x\right)=\frac{2x^3-x^2+1}{x^2+1}f(x)=x2+12x3−x2+1
Below
Coincides
Determine the relative position of the asymptote y=3y=3y=3 to the curve of f(x)=2x2−x+3x2+1f\left(x\right)=\frac{2x^2-x+3}{x^2+1}f(x)=x2+12x2−x+3
Determine the relative position of the asymptote y=−2x+3y=-2x+3y=−2x+3 to the curve of f(x)=4x2−2x+3xf\left(x\right)=\frac{4x^2-2x+3}{x}f(x)=x4x2−2x+3
Above
Determine the relative position of the asymptote y=−3xy=-3xy=−3x to the curve of f(x)=3x3+4x−2xf\left(x\right)=\frac{3x^3+4x-2}{x}f(x)=x3x3+4x−2
Find the relative position of the asymptote y=5y=5y=5 to the curve of f(x)=5x2+2x+1x2+3f\left(x\right)=\frac{5x^2+2x+1}{x^2+3}f(x)=x2+35x2+2x+1
Find the relative position of the asymptote y=4y=4y=4 to the curve of f(x)=4x3−2x2+1x2+1f\left(x\right)=\frac{4x^3-2x^2+1}{x^{2+1}}f(x)=x2+14x3−2x2+1
Determine the relative position of the asymptote y=2xy=2xy=2x to the curve of f(x)=3x2+2x−1xf\left(x\right)=\frac{3x^2+2x-1}{x}f(x)=x3x2+2x−1
Find the relative position of the asymptote y=−1y=-1y=−1 to the curve of f(x)=x3−2x+3x+1f\left(x\right)=\frac{x^3-2x+3}{x+1}f(x)=x+1x3−2x+3
Determine the relative position of the asymptote y=−xy=-xy=−x to the curve of f(x)=2x2+x−3x+1f\left(x\right)=\frac{2x^2+x-3}{x+1}f(x)=x+12x2+x−3
Find the relative position of the asymptote y=2y=2y=2 to the curve of f(x)=3x2+1x−1f\left(x\right)=\frac{3x^2+1}{x-1}f(x)=x−13x2+1
It is done.