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Find the horizontal or oblique asymptote of f(x) = 2 x squared plus 5 x plus 6, all over x plus 1

A) y = −1
B) y = −2
C) y = 2x + 3
D) y = −2x + 5

Respuesta :

ghanami
hello : 
f(x) = (2x²+5x+6)/(x+1)
Answer (B) because lim  (2x²+5x+6)/(x+1) - (2x+3) = 0 
x in infinity + or -

Answer:

y=2x+3

Step-by-step explanation:

[tex]f(x)=\frac{2x^2+5x+6}{x+1}[/tex]

Degree of numerator is 2 and degree of denominator is 1

degree of numerator is greater than the degree of denominator by 1, so there will be oblique asymptote

to find oblique asymptote we use long division

Divide the first term by first term of divisor and put is at the top

Multiply it by x+1 and put it down, repeat the process till we get remainder.

                                              [tex]2x+3[/tex]

                             --------------------------------------

[tex]x+1[/tex]                    [tex]2x^2+5x+6[/tex]

                                        [tex]2x^2+2x[/tex]

                         ------------------------------------------------(subtract)

                                                   [tex]3x+6[/tex]

                                                     [tex]3x+3[/tex]

                               -----------------------------------------------

                                                                  3

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