Word Problem 3 Help !!!

[merikukri]

3.Create a rational function that has the following characteristics: crosses the x-axis at F; touches the x-axis at -L ; has one vertical asymptote at x=-L-1 and another at x=F+1 ; and has one horizontal asymptote, y=F . Give the formula and a complete graph

Where F = 5, L =4

 

[Gene]

3. In order, the requirements are:
(x-5); (x+4)^2; 1/(x-5); 1/(x-6); 5/(x-6).

5(x-5)(x+4)^2/((x-5)(x-6)^2)

 

[merikukri]

3. In order, the requirements are:
(x-5); (x+4)^2; 1/(x-5); 1/(x-6); 5/(x-6).

5(x-5)(x+4)^2/((x-5)(x-6)^2)

Hey Gene thanks for help, but can u explain how you got that.
Thanks for putting time in doing work, I really appreciate it.

How did u get (x+4)^2 in numerator and (x-6)^2 in denominator ?

Regards

 

[stapel]

How did u get (x+4)^2 in numerator...?

How does one get a zero of a rational function? What must be zero for the entire fraction to be zero? What form must that zero take for the graph to only touch the axis, rather than cross it? (This question requires that you use your knowledge gained from graphing various functions.)

How did u get...(x-6)^2 in denominator ?

One of the factors in the denominator had to be squared, in order to get the comparative degrees right. So think about what sort of conditions you need to have to have an horizontal asymptote, and to have that asymptote be something other than the x-axis.

Eliz.

 

[merikukri]

How did u get (x+4)^2 in numerator...?

How does one get a zero of a rational function? What must be zero for the entire fraction to be zero? What form must that zero take for the graph to only touch the axis, rather than cross it? (This question requires that you use your knowledge gained from graphing various functions.)

How did u get...(x-6)^2 in denominator ?

One of the factors in the denominator had to be squared, in order to get the comparative degrees right. So think about what sort of conditions you need to have to have an horizontal asymptote, and to have that asymptote be something other than the x-axis.

Eliz.

Thanks Eliz, this is very helpful.