Asked by big145
Write an equation to represent a rational function with the following conditions:
- reciprocal of a quadratic function
- asymptotes with the equations x = -5, x= 4/3, and y = 0
- y < 0 whenever x < -5
- y-intercept of 1.5
- reciprocal of a quadratic function
- asymptotes with the equations x = -5, x= 4/3, and y = 0
- y < 0 whenever x < -5
- y-intercept of 1.5
Answers
Answered by
oobleck
the roots are -5 and 4/3, so start with
y = a/((x+5)(3x-4))
This also has y=0 as the horizontal asymptote
When x < -5, both factors are negative, so a must be negative
Now, using the point (0,3/2), we get
3/2 = a/(-5 * -19)
a = 142.5
But that is positive. So we need to make that denominator positive, without changing the asymptotes. One way is to square a factor.
y = a/((x+5)^2 (3x-4))
Now at x=0,
3/2 = a/(25 * -4)
a = -150
and so
y = -150/((x+5)^2 (3x-4)) = -150/(3x^3+26x^2+35x-100)
y = a/((x+5)(3x-4))
This also has y=0 as the horizontal asymptote
When x < -5, both factors are negative, so a must be negative
Now, using the point (0,3/2), we get
3/2 = a/(-5 * -19)
a = 142.5
But that is positive. So we need to make that denominator positive, without changing the asymptotes. One way is to square a factor.
y = a/((x+5)^2 (3x-4))
Now at x=0,
3/2 = a/(25 * -4)
a = -150
and so
y = -150/((x+5)^2 (3x-4)) = -150/(3x^3+26x^2+35x-100)
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