Question

Find the horizontal or oblique asymptote of f(x) = negative 2 x squared plus 3 x plus 6, all over x plus 2 .

Answer 1

$f(x) = \frac{-2x^{2} + 3x + 6}{x + 2}$

                  -2x + 7
x + 2|-2x² + 3x + 6
         -2x² -  4x
                   7x + 6
                  -7x + 14 
                   14x - 8

$f(x) = \frac{-2x^{2} + 3x + 6}{x + 2} = -\frac{2x^{2} + 3x + 6}{x + 2} = -2x + 7 + \frac{14x - 8}{x + 2}$

y = -2x + 7

The answer is A.

Answer 2

Answer:  First option is correct.

Step-by-step explanation:

Since we have given that

$f(x)=\frac{-2x^2+3x+6}{x+2}$

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.

$\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}-2x^2+3x+6\mathrm{\:and\:the\:divisor\:}x+2\mathrm{\::\:}\frac{-2x^2}{x}=-2x\\\\\mathrm{Quotient}=-2x\\\\\mathrm{Multiply\:}x+2\mathrm{\:by\:}-2x:\:-2x^2-4x\\\\\mathrm{Subtract\:}-2x^2-4x\mathrm{\:from\:}-2x^2+3x+6\mathrm{\:to\:get\:new\:remainder}\\\\\mathrm{Remainder}=7x+6\\\\\mathrm{Remainder}=7x+6\\\\=-2x+\frac{7x+6}{x+2}$

And,

$\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}7x+6\mathrm{\:and\:the\:divisor\:}x+2\mathrm{\::\:}\frac{7x}$${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}$

At last, we get,

$-2x+7-\frac{8}{x+2}$

Hence, the oblique asymptote of f(x) is

y=-2x+7

Therefore, First option is correct.