Answer:
slant asymptote is f(x) = x - 1
see graph below
Explanation:
Given:
[tex]f(x)\text{ = }\frac{-x^2\text{ - x + 3}}{-x-2}[/tex]
To find:
The slant asymptote and graph it
[tex]\begin{gathered} Using\text{ long division:} \\ Quotient\text{ = x - 1} \\ remainder\text{ = 1} \\ \frac{-x^2\text{ - x + 3}}{-x-2}\text{ = \lparen x - 1\rparen + }\frac{1}{-x-2} \\ taking\text{ the limit, as x tends to }\infty,\text{ }\frac{1}{-x-2}\text{ tends to zero} \\ \\ we\text{ would be left with = x - 1} \\ The\text{ slant asymptote, f\lparen x\rparen = x - 1} \end{gathered}[/tex]
Graphing the equation on the same coordinate as the function:
