document.write( "Question 1086481: If f(x)=(x^5-6x^4+7x^3+6x^2-8x)/(x^2-2x)\r
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document.write( "(a) Find all intercept(s) if any.\r
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document.write( "(b) Find all asymptote(s), hole(s), gap(s),jump(s) if any.\r
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document.write( "(c) Graph.\r
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document.write( "(d) Give the domain and range. \n" );
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Algebra.Com's Answer #700715 by Boreal(15235) You can put this solution on YOUR website! f(x)=(x^5-6x^4+7x^3+6x^2-8x)/(x^2-2x) \n" ); document.write( "factor out an x(x^4-6x^3+7x^2+6x-8)/x(x-2) \n" ); document.write( "x cancel, \n" ); document.write( "x can't be 0, but they cancel, so there is a hole at x=0. \n" ); document.write( "suspect 2 is a root, so (x-2) a factor \n" ); document.write( "2/1===-6===7===6===-8 \n" ); document.write( "=1====-4=-1====4===0 \n" ); document.write( "it is \n" ); document.write( "the other factor is (x^3-4x^2-x+4) \n" ); document.write( "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) \n" ); document.write( "1/1====-4===-1====4 \n" ); document.write( "==1====-3===-4===0 \n" ); document.write( "(x-1) is a root. \n" ); document.write( "That leaves x^2-3x-4, and that factors into (x-4)(x+1) \n" ); document.write( "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. \n" ); document.write( "Factored, the function is \n" ); document.write( "(x+1)(x-1)(x-2)(x-4)*x/x(x-2) \n" ); document.write( "That becomes (x+1)(x-1)(x-4). \n" ); document.write( "Domain is all x except x=0,2 \n" ); document.write( "range is all real numbers \n" ); document.write( "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. \n" ); document.write( " |