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# Write an equation for a rational function with: Vertical asymptotes at x = 4 and x = 6 x intercepts at x = -4 and x = 2 Horizontal asymptote at y = 9

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Write an equation for a rational function with: Vertical asymptotes at x = 4 and x = 6 x intercepts at x = -4 and x = 2 Horizontal asymptote at y = 9

y =_______

Guest Jun 11, 2015

#1
+92857
+10

Write an equation for a rational function with: Vertical asymptotes at x = 4 and x = 6 x intercepts at x = -4 and x = 2 Horizontal asymptote at y = 9

y = [9(x+4) (x -2)] / [(x -4) (x -6)]   =  [9x^2 + 18x - 72]  / [x^2 - 10x + 24]

Here's a graph.........https://www.desmos.com/calculator/iowejkz3ld

CPhill  Jun 11, 2015
#1
+92857
+10

Write an equation for a rational function with: Vertical asymptotes at x = 4 and x = 6 x intercepts at x = -4 and x = 2 Horizontal asymptote at y = 9

y = [9(x+4) (x -2)] / [(x -4) (x -6)]   =  [9x^2 + 18x - 72]  / [x^2 - 10x + 24]

Here's a graph.........https://www.desmos.com/calculator/iowejkz3ld

CPhill  Jun 11, 2015
#2
+94183
+5

Thanks Chris but I am a little confused.

Write an equation for a rational function with: Vertical asymptotes at x = 4 and x = 6 x intercepts at x = -4 and x = 2 Horizontal asymptote at y = 9

Since the roots are x=-4 and x=2  The numerator must contain  (x+4)(x-2)

And since  x=4 and x=6 are  aymptotes the denominator must contain     (x-4)(x-6)

BUT i don't understand the logic behind the horizontal asymptote at y=9

**I have added this to our reference material thread so if it is expanded upon that would be really good

Melody  Jun 11, 2015
#3
+92857
+5

Remember, Melody, that when we have a "same over same" situation, the horizontal asymptote will just be the ratio of the coefficients on  the highest powers of the variable x.....this will be (9x^2)/(x^2), or just, 9.

Seen another way, just take the limit of the function as x approaches infinity.........

So ... lim as x →∞   [9x^2 + 18x - 72]  / [x^2 - 10x + 24].........divide all terms by x^2........18x, -72, -10x and 24  will approach 0 ..... and  we are left with  9x^2 / x^2    which approaches 9 as x →∞

CPhill  Jun 11, 2015
#4
+20687
+5

BUT i don't understand the logic behind the horizontal asymptote at y=9

$$\small{\text{ \lim \limits_{x \rightarrow \infty } \dfrac{(x+4)(x-2)} {(x-4)(x-6)} = \lim \limits_{x \rightarrow \infty } \dfrac{x^2+2x-8} {x^2-10x+24} = \lim \limits_{x \rightarrow \infty } \dfrac { \dfrac{x^2}{x^2}+\dfrac{2x}{x^2}-\dfrac{8}{x^2}} { \dfrac{x^2}{x^2}-\dfrac{10x}{x^2}+\dfrac{24}{x^2}} = \lim \limits_{x \rightarrow \infty } \dfrac { 1 + \dfrac{2}{x}-\dfrac{8}{x^2}} { 1 - \dfrac{10}{x}+\dfrac{24}{x^2}} = \dfrac { 1 } { 1 }=1 }}\\\\$$

$$\small{\text{ \lim \limits_{x \rightarrow \infty } 9\cdot \dfrac{(x+4)(x-2)} {(x-4)(x-6)} = \lim \limits_{x \rightarrow \infty } 9\cdot \dfrac{x^2+2x-8} {x^2-10x+24} = \lim \limits_{x \rightarrow \infty } 9\cdot \dfrac { \dfrac{x^2}{x^2}+\dfrac{2x}{x^2}-\dfrac{8}{x^2}} { \dfrac{x^2}{x^2}-\dfrac{10x}{x^2}+\dfrac{24}{x^2}} = \lim \limits_{x \rightarrow \infty } 9\cdot \dfrac { 1 + \dfrac{2}{x}-\dfrac{8}{x^2}} { 1 - \dfrac{10}{x}+\dfrac{24}{x^2}} = 9\cdot \dfrac { 1 } { 1 }=9 }}\\\\$$

heureka  Jun 11, 2015
#5
+94183
0

Thanks Chris and Heureka,

That makes sense

Melody  Jun 11, 2015