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# Find the horizontal asymptote of f(x) = 2 ((x+4)(5 x-1))/((8-x)(7 x + 2)) y =

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Find the horizontal asymptote of f(x) = 2 ((x+4)(5 x-1))/((8-x)(7 x + 2))

y =______

Jun 11, 2015

#3
+20850
+5

Find the horizontal asymptote of  f(x) = 2 ((x+4)(5 x-1))/((8-x)(7 x + 2))

$$\small{\text{ \lim \limits_{x \rightarrow \infty } 2\cdot \dfrac{(x+4)(5x-1)}{(8-x)(7x+2)} =\lim \limits_{x \rightarrow \infty } 2\cdot \dfrac{5x^2+19x-4}{-7x^2+54x+16} =\lim \limits_{x \rightarrow \infty } 2\cdot \dfrac{ \dfrac{5x^2}{x^2}+\dfrac{19x}{x^2}-\dfrac{4}{x^2}}{ \dfrac{-7x^2}{x^2}+\dfrac{54x}{x^2}+\dfrac{16}{x^2}} =\lim \limits_{x \rightarrow \infty } 2\cdot \dfrac{ 5 + \dfrac{19}{x}-\dfrac{4}{x^2}}{ -7 + \dfrac{54}{x}+\dfrac{16}{x^2}} = 2\cdot \dfrac{ 5 }{ -7 }=-\dfrac{10}{7} }}$$

$$\rm{horizontal~ Asymptote:~} y =-\dfrac{10}{7} = -1.42857142857$$

Jun 11, 2015

#1
+95361
+5

Well lets see if I can use Chris's shortcut that he just showed me

asymptote at

$$\\y=\frac{2*5x^2}{-7x^2}\\\\ y=\frac{-10}{7}\\\\$$

Is that right?   I know I could check myself but I am being lazy :/

Jun 11, 2015
#2
+94607
+5

Yep, Melody.....correct........see the graph here......https://www.desmos.com/calculator/lbfbn2mv6p

Jun 11, 2015
#3
+20850
+5

Find the horizontal asymptote of  f(x) = 2 ((x+4)(5 x-1))/((8-x)(7 x + 2))

$$\small{\text{ \lim \limits_{x \rightarrow \infty } 2\cdot \dfrac{(x+4)(5x-1)}{(8-x)(7x+2)} =\lim \limits_{x \rightarrow \infty } 2\cdot \dfrac{5x^2+19x-4}{-7x^2+54x+16} =\lim \limits_{x \rightarrow \infty } 2\cdot \dfrac{ \dfrac{5x^2}{x^2}+\dfrac{19x}{x^2}-\dfrac{4}{x^2}}{ \dfrac{-7x^2}{x^2}+\dfrac{54x}{x^2}+\dfrac{16}{x^2}} =\lim \limits_{x \rightarrow \infty } 2\cdot \dfrac{ 5 + \dfrac{19}{x}-\dfrac{4}{x^2}}{ -7 + \dfrac{54}{x}+\dfrac{16}{x^2}} = 2\cdot \dfrac{ 5 }{ -7 }=-\dfrac{10}{7} }}$$

$$\rm{horizontal~ Asymptote:~} y =-\dfrac{10}{7} = -1.42857142857$$

heureka Jun 11, 2015
#4
+95361
0

Thanks guys

Jun 11, 2015