1.

Find the asymptote of the function ` y = (2x^(2) + 3x + 1)/ x` if any.

Answer» Vertical asymptote:
Since denominator is x, vertical asymptote is x = 0.
Oblique asymptote:
Let the oblique asymptote by y = mx + n. Therefore,
`underset(x to infty)"lim"[(2x^(2)+3x+1)/x -(mx + n)] = 0`
` rArrunderset(x to infty)"lim"[(2x^(2)+3x+1-(mx+n)x)/x ] =0`
` rArr underset(x to infty)"lim"[[(2-m)x^(2)+(3-n)x+1)/x]=0`
Clearly, 2 - m = 0
` rArr underset(x to infty)"lim" [((3-n)x+1)/x]=0`
` rArr underset(x to infty)"lim"[(3-n)+1/x]=0`
` rArr 3- n = 0`
` rArr n = 3`
Thus, the oblique asymptote is y = 2x + 3.
Alternative method:
We have ` (2x^(2)+3x+1)/x = 2x+3+1/x`
Here quotient is 2x+3, hence the oblique asymptote is y = 2x +3.


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