SOLUTION: Find an equation of a rational function 𝑓 that satisfies the given conditions:
Vertical asymptotes: 𝑥 = −5, 𝑥 = 2
Horizontal asymptotes: 𝑦 = −2
𝑥-intercepts:
Algebra.Com
Question 1161728: Find an equation of a rational function 𝑓 that satisfies the given conditions:
Vertical asymptotes: 𝑥 = −5, 𝑥 = 2
Horizontal asymptotes: 𝑦 = −2
𝑥-intercepts: 𝑥 = −6, 𝑥 = 4
𝑓(−2) = −4
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Vertical asymptotes occur at values of the independent variable that make the denominator equal to zero. Therefore, if there are vertical asymptotes at 
and 
, then the denominator polynomial must, at a minimum, have factors of 
and 
.
If there is a non-zero horizontal asymptote, the degrees of the numerator and denominator polynomials are equal, and the asymptote occurs at 
where 
is the lead coefficient of the numerator polynomial and 
is the lead coefficient of the denominator polynomial. Hence, the lead coefficients of the numerator and denominator polynomials in the desired rational function must be in the ratio 
.
-intercepts occur at values of that make the numerator equal to zero. Hence, the numerator must, at a minimum, have factors of 
and 
.
Putting together what we know so far:
(x\ -\ 4)}{(x\ +\ 5)(x\ -\ 2)})
Checking the final condition:
\ =\ \frac{-2(-2\ +\ 6)(-2\ -\ 4)}{(-2\ +\ 5)(-2\ -\ 2)}\ =\ \frac{-2\,\cdot\,4\,\cdot\,-6}{3\,\cdot\,-4}\ =\ \frac{48}{-12}\ =\ -4)
So
\ =\ \frac{-2(x\ +\ 6)(x\ -\ 4)}{(x\ +\ 5)(x\ -\ 2)})
Meets all of the stated conditions.
John
My calculator said it, I believe it, that settles it
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