SOLUTION: If f(x)=(x^5-6x^4+7x^3+6x^2-8x)/(x^2-2x)
(a) Find all intercept(s) if any.
(b) Find all asymptote(s), hole(s), gap(s),jump(s) if any.
(c) Graph.
(d) Give the domain and
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Question 1086481: If f(x)=(x^5-6x^4+7x^3+6x^2-8x)/(x^2-2x)
(a) Find all intercept(s) if any.
(b) Find all asymptote(s), hole(s), gap(s),jump(s) if any.
(c) Graph.
(d) Give the domain and range.
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
f(x)=(x^5-6x^4+7x^3+6x^2-8x)/(x^2-2x)
factor out an x(x^4-6x^3+7x^2+6x-8)/x(x-2)
x cancel,
x can't be 0, but they cancel, so there is a hole at x=0.
suspect 2 is a root, so (x-2) a factor
2/1===-6===7===6===-8
=1====-4=-1====4===0
it is
the other factor is (x^3-4x^2-x+4)
if x=1, that will be a root by inspection (just put 1 in for x and see if the coefficients add up to 0. They do)
1/1====-4===-1====4
==1====-3===-4===0
(x-1) is a root.
That leaves x^2-3x-4, and that factors into (x-4)(x+1)
The roots are -1,0,1,2,4. But the denominator factors into x(x-2), so 0 and 2 are holes and -1,1,4 are roots.
Factored, the function is
(x+1)(x-1)(x-2)(x-4)*x/x(x-2)
That becomes (x+1)(x-1)(x-4).
Domain is all x except x=0,2
range is all real numbers
Polynomial division gives a quotient of x^3-4x^2-4x+4, which has no remainder. The original function has holes at x=0 and x=2; the slant asymptote does not.
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