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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

Best Answer 

 #1
avatar+78755 
+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
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5+0 Answers

 #1
avatar+78755 
+10
Best Answer

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
avatar+91051 
+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
avatar+78755 
+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
avatar+18715 
+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
avatar+91051 
0

Thanks Chris and Heureka,

That makes sense  

Melody  Jun 11, 2015

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