# Find the horizontal or oblique asymptote of f (x) = 2x^{2} + 5x + 6, all over x plus 1.

**Solution:**

Given f (x) = 2x^{2} + 5x + 6 all over x+1

f (x) = (2x^{2} + 5x + 6)/x+1

The horizontal asymptote is 0 as the degree of the numerator > the degree of the denominator.(2 >1)

Denominator =0 when x= -1

So, vertical asymptote is at x=-1

We get the oblique aysmptote when the degree of the numerator exceeds that of the denominator by 1.

f (x) =(2x^{2} + 2x + 3x+ 3 + 3)/x+1

= (2x(x+1) + 3(x +1) + 3)/x+1

= [(2x+3)(x+1) + 3]/(x+1)

= (2x+3) + [3/(x+1)]

Here oblique asymptote is 2x+3

## Find the horizontal or oblique asymptote of f (x) = 2x^{2} + 5x + 6, all over x plus 1.

**Summary: **

The horizontal or oblique asymptote of f (x) = 2x^{2} + 5x + 6, all over x plus 1 is as shown above.

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