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**+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**+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**+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

Thanks Chris but I am a little confused.

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**+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**+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