Answer: First option is correct.
Step-by-step explanation:
Since we have given that
[tex]f(x)=\frac{-2x^2+3x+6}{x+2}[/tex]
We need to find the horizontal or oblique asymptote .
Since the degree of numerator is more than the degree of denominator.
So, it has oblique asymptote.
[tex]\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}-2x^2+3x+6\mathrm{\:and\:the\:divisor\:}x+2\mathrm{\::\:}\frac{-2x^2}{x}=-2x\\\\\mathrm{Quotient}=-2x\\\\\mathrm{Multiply\:}x+2\mathrm{\:by\:}-2x:\:-2x^2-4x\\\\\mathrm{Subtract\:}-2x^2-4x\mathrm{\:from\:}-2x^2+3x+6\mathrm{\:to\:get\:new\:remainder}\\\\\mathrm{Remainder}=7x+6\\\\\mathrm{Remainder}=7x+6\\\\=-2x+\frac{7x+6}{x+2}[/tex]
And,
[tex]\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}7x+6\mathrm{\:and\:the\:divisor\:}x+2\mathrm{\::\:}\frac{7x}[/tex][tex]{x}=7\\\\\mathrm{Quotient}=7\\\\\mathrm{Multiply\:}x+2\mathrm{\:by\:}7:\:7x+14\\\\\mathrm{Subtract\:}7x+14\mathrm{\:from\:}7x+6\mathrm{\:to\:get\:new\:remainder}\\\\\mathrm{Remainder}=-8\\\\Therefore,\\\\\frac{7x+6}{x+2}=7+\frac{-8}{x+2}[/tex]
At last, we get,
[tex]-2x+7-\frac{8}{x+2}[/tex]
Hence, the oblique asymptote of f(x) is
y=-2x+7
Therefore, First option is correct.
