Asked by Gideon
Please help
find an equation of a rational function given the following properties
Vertical asymptote: x=-2
Slant Asymptote: y=x-5
domain : (-inf, -2)U(-2,-1)U(-1,inf)
x intercept: x=3 Y intercept: y=12
hole:(1,4)
find an equation of a rational function given the following properties
Vertical asymptote: x=-2
Slant Asymptote: y=x-5
domain : (-inf, -2)U(-2,-1)U(-1,inf)
x intercept: x=3 Y intercept: y=12
hole:(1,4)
Answers
Answered by
oobleck
Vertical asymptote: x=-2
y = a/(x+2)
x intercept: x=3
y = a(x-3)/(x+2)
hole:(1,4)
y = a(x-3)(x-1) / (x-1)(x+2)
domain : (-inf, -2)U(-2,-1)U(-1,inf)
you also need a hole at x = -1
y = a(x-3)(x-1)(x+1) / (x+1)(x-1)(x+2)
Slant Asymptote: y=x-5
You need the above to get small as x gets big, but without adding any more holes or asymptotes
y = x-5 + (a(x-3)(x-1)(x+1)) / ((x+1)(x-1)(x+2)(x^2+1))
Y intercept: y=12
solve for a in y(0)=12 gives a = -34/3
y = x-5 + ((-34/3)(x-3)(x-1)(x+1)) / ((x+1)(x-1)(x+2)(x^2+1))
now collect terms to make that a single rational expression
y = ((3x^4-9x^3-27x^2-43x+72)(x-1)(x+1)) / ((x+1)(x-1)(x+2)(x^2+1))
y = a/(x+2)
x intercept: x=3
y = a(x-3)/(x+2)
hole:(1,4)
y = a(x-3)(x-1) / (x-1)(x+2)
domain : (-inf, -2)U(-2,-1)U(-1,inf)
you also need a hole at x = -1
y = a(x-3)(x-1)(x+1) / (x+1)(x-1)(x+2)
Slant Asymptote: y=x-5
You need the above to get small as x gets big, but without adding any more holes or asymptotes
y = x-5 + (a(x-3)(x-1)(x+1)) / ((x+1)(x-1)(x+2)(x^2+1))
Y intercept: y=12
solve for a in y(0)=12 gives a = -34/3
y = x-5 + ((-34/3)(x-3)(x-1)(x+1)) / ((x+1)(x-1)(x+2)(x^2+1))
now collect terms to make that a single rational expression
y = ((3x^4-9x^3-27x^2-43x+72)(x-1)(x+1)) / ((x+1)(x-1)(x+2)(x^2+1))
Answered by
Gideon
so the denominator going to be (x+1)(x-1)(x+2)(x^2+1)
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