Find an equation of a rational function

  • Thread starter Thread starter DDRchick
  • Start date Start date
  • Tags Tags
    Function Rational
AI Thread Summary
To find the equation of a rational function f(x) with specified conditions, the vertical asymptotes at x = -5 and x = 0 lead to factors of (x + 5) and x in the denominator. The x-intercept at 7 results in the factor (x - 7) in the numerator. The horizontal asymptote at y = 0 is achieved because the degree of the denominator is higher than that of the numerator. A constant 'a' is introduced to satisfy the condition f(1) = 4, allowing for the final equation f(x) = -4(x - 7)/[x(x + 5)]. Understanding the placement of 'a' and the factors clarifies how the function meets all given conditions.
DDRchick
Messages
27
Reaction score
0
Find an equation of a rational function f(x)that satisfies the conditions. We're allowed to use a calculator, by the way. (:
Okay, conditions:
vertical asymptote: x=-5, x=0
Horizontal asymptote: y=0
x-intercept=7; f(1)=4




On the test i had no idea how to do it, but after seeing her key,I somewhat understand. I looked in the textbook and was able to see where many things came from.
a(x-7)/(x+5)(x)
for one, i have no idea where the a and the x came from. I do, however, understand that the (x-7) is from the x-int. and that the (x+5) is from the v.a.
4=a(1-7)/(1+5)(x) again, no idea where the a and x came from. I see that 4=f(1) which means that y=4 and x=1.
4=a(-6)/(6)(1)
I understand she plugged in that last x finally with a 1 and simplified everything else.
4=-a, a=-4
f(x)=-4(x-7)/x(x+5)
I understand the final equation. I'd just love to know were she got that a and the x (by iself on the denominator) from! Thanks (:
 
Physics news on Phys.org
The vertical asymptotes of x = -5 and 0 gave the factors of (x+5) and (x-0) in the denominator. The x intercept of 7 gives the (x-7) in the numerator.

That's where the fraction \frac{x-7}{(x+5)(x)} comes from. The second degree in the denominator vs. first degree in the numberator gives the horizontal asymptote of 0 for free. You have one condition left and only need to note that multiplying the fraction by a constant a doesn't change any of the above features and let's you get the last constraint.
 
Ohhh okay thanks so much! :]
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top