Question

Write a rational function that has the following properties:
• horizontal asymptote y = −3
• vertical asymptotes x = 0 , x = 5 x = −2
• its only -intercept is x = 2

Answer 1

Step-by-step explanation:

Rational function is 2x2−22x+565x2−20x−105

Explanation:

A zero at x=4 means we have (x−4) as a factor in numerator;

a hole at x=7 means, we have x−7 a factor both in numerator as well as denominator;

a vertical asymptote at x=−3 means x+3 a factor in denominator only

a horizontal asymptote at y=25 means highesr degrees in both numerator and denominator are equal and their coefficients are in ratio of 2:5

Hence desired rational function is 2(x−4)(x−7)5(x−7)(x+3)

i.e. 2x2−22x+565x2−20x−105

See its graph down below. Observe vertical asymptote x=−3 and horizontal asymptote y=25. We have a zero at x=4 as functtion passes through (4,0). Hole is not seen as x−7 cancels out, but we know, the function is not definedat this point.

graph{(2x^2-22x+56)/(5x^2-20x-105) [-10.67, 9.33, -4.4, 5.6]} 1

A rational function, R(x) has the following characteristics:

a vertical asymptote at x = 3,

a horizontal asymptote at y = 2,

and a hole at (2, −2).

Sketch the function and determine what it could be using the following steps:

Put in the factor that would account for the vertical asymptote at x = 3.

Add in the factors that would account for a hole at x = 2.

l

Determine what must be true about the numerator and denominator for there to be a horizontal asymptote at y = 2.

Add the factors that would account for the horizontal asymptote at y = 2.

x-intercept of 3 ---> needs a multiplier of x-3

vertical asymtote of x = -2 ----> needs a divisor of x+2

so far y = (x-3)/(x+2)

horizontal asymptote of y = 2

lets slap a 2 infront

y = 2(x-3)/(x+2) , as x gets huge y ---> 2 , ( try x = 100,000 )

we want y-intercept of -3, (when x = 0)

We lucked out, since if x = 0 , y = -6/2 = -3

y = 2(x-3)/(x+2)