+0

# 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

0
412
5

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
+80922
+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
Sort:

#1
+80922
+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
+91436
+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
+80922
+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
+18827
+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
+91436
0

Thanks Chris and Heureka,

That makes sense

Melody  Jun 11, 2015

### 18 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details