Question

What is the oblique asymptote of the function f(x) = the quantity of x squared plus 4 x plus 5, all over x plus 3?

Answer 1

f(x)=(x²+4x+5)/(x+3)

to find the oblique asymptote, you just proceed to the division of::

(x²+4x+5) by (x+3) & the quotient will be the oblique linear function which will be the asymptote:
The quotient of the division is x+1, you ignore the rest if any
& the oblique asymptote is f(x) = x+1

Answer 2

Answer: Oblique asymptote of the function is y=x+1.

Explanation:

Since we have given that

`f(x)=(x^2+4x+5)/(x+3)`

So, it is in the form of
`(p(x))/(q(x))`

So, first we need to find the simplest form of numerator and denominator by division.

`(x^2+4x+5)/(x+3)\\\\\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}x^2+4x+5\mathrm{\:and\:the\:divisor\:}x+3\mathrm{\::\:}\\\\(x^2)/(x)=x\\\\\mathrm{Quotient}=x\\\\\mathrm{Multiply\:}x+3\mathrm{\:by\:}x:\:x^2+3x\\\\\mathrm{Subtract\:}x^2+3x\mathrm{\:from\:}x^2+4x+5\mathrm{\:to\:get\:new\:remainder}\\\\\mathrm{Remainder}=x+5\\\\\mathrm{Therefore}\\\\(x^2+4x+5)/(x+3)=x+(x+5)/(x+3)\\\\`

`\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}x+5\mathrm{\:and\:the\:divisor\:}x+3\mathrm{\::\:}(x)/(x)=1\\\\\mathrm{Quotient}=1\\\\\mathrm{Multiply\:}x+3\mathrm{\:by\:}1:\:x+3\\\\\mathrm{Subtract\:}x+3\mathrm{\:from\:}x+5\mathrm{\:to\:get\:new\:remainder}\\\\\mathrm{Remainder}=2\\\\Therefore\\\\(x+5)/(x+3)=1+(2)/(x+3)`

Hence,

`(x^2+4x+5)/(x+3)=x+1+(x+5)/(x+3)\\`

So, Oblique asymptote of the function is y=x+1.