Answer: Oblique asymptote of the function is y=x+1.
Step-by-step explanation:
Since we have given that
[tex]f(x)=\frac{x^2+4x+5}{x+3}[/tex]
So, it is in the form of [tex]\frac{p(x)}{q(x)}[/tex]
So, first we need to find the simplest form of numerator and denominator by division.
[tex]\frac{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{\::\:}\\\\\frac{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}\\\\\frac{x^2+4x+5}{x+3}=x+\frac{x+5}{x+3}\\\\[/tex]
[tex]\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}x+5\mathrm{\:and\:the\:divisor\:}x+3\mathrm{\::\:}\frac{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\\\\\frac{x+5}{x+3}=1+\frac{2}{x+3}[/tex]
Hence,
[tex]\frac{x^2+4x+5}{x+3}=x+1+\frac{x+5}{x+3}\\[/tex]
So, Oblique asymptote of the function is y=x+1.
