document.write( "Question 815092: Hi I've been trying to find out what to do and how to do and I'm stuck so I'm hoping you can help me
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\n" ); document.write( "What is f of x equal x square plus seven x plus ten over x square subtract seven x subtract eighteen \r
\n" ); document.write( "\n" ); document.write( "It looks like F(x) = x^2+7x+10
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Algebra.Com's Answer #490711 by Edwin McCravy(20056)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "f(x) = \"%28x%5E2%2B7x%2B10%29%2F%28x%5E2-7x-18%29\"\r\n" );
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document.write( "Since the degree of the numerator and denominator are both 2,\r\n" );
document.write( "the horizontal asymptote has the equation\r\n" );
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document.write( "y = \"%28LEADING_COEFFICIENT_OF_NUMERATOR%29%2F%28LEADING_COEFFICIENT_OF_DENOMINATOR%29\"\r\n" );
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document.write( "y = \"1%2F1\"\r\n" );
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document.write( "y = 1\r\n" );
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document.write( "Let's draw the horizontal asymptote y = 1 (in green) \r\n" );
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document.write( "This function f(x) is not defined when the denominator = 0.\r\n" );
document.write( "It either has a vertical asymptote there or else it\r\n" );
document.write( "has a hole in the curve.\r\n" );
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document.write( "We set the denominator = 0.\r\n" );
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document.write( "  x²-7x-18 = 0\r\n" );
document.write( "(x-9)(x+2) = 0\r\n" );
document.write( "x-9=0;  x+2=0\r\n" );
document.write( "  x=9;    x=-2\r\n" );
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document.write( "So the function is not defined at either x=9 or x=-2.\r\n" );
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document.write( "The domain of f(x) is (-oo,-2)U(-2,9)U(9,oo)\r\n" );
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document.write( "Next we find out if there is a vertical asymptote or\r\n" );
document.write( "a \"hole in the curve\" (removable discontinuity)\r\n" );
document.write( "at x=9 and x=-2\r\n" );
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document.write( "Factor the numerator x²+7x+10 as (x+5)(x+2)\r\n" );
document.write( "We have already factored the denominator x²-7x-18 as (x-9)(x+2).\r\n" );
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document.write( "f(x) = \"%28%28x%2B5%29%28x%2B2%29%2F%28x-9%29%28x%2B2%29%29\"\r\n" );
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document.write( "Now we may ONLY cancel the (x+2)'s if we specify that x is not\r\n" );
document.write( "equal to -2, for f(x) is not defined at x=-2 or at x=9.\r\n" );
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document.write( "But the fact that we have a factor (x-2) in the numerator and \r\n" );
document.write( "also an (x-2) factor in the denominator tells us there is a\r\n" );
document.write( "removable discontinuity at x=-2.  And since there is no (x-9)\r\n" );
document.write( "factor in the numerator tells us that there is a vertical\r\n" );
document.write( "asymptote at x=9.\r\n" );
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document.write( "So we draw the vertical asymptote (also in green) which has \r\n" );
document.write( "the equation x=9\r\n" );
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document.write( "Now if we cancel the (x+2)'s we will get a function that we will\r\n" );
document.write( "call g(x).  g(x) is exactly like f(x) except it will have a value at \r\n" );
document.write( "x=2 whereas f(x) does not have a value there.\r\n" );
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document.write( "So the graph that removes the dicontinuity (plugs up the hole) is\r\n" );
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document.write( "g(x) = \"%28x%2B5%29%2F%28x-9%29\" has an asymptote at x=9\r\n" );
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document.write( "Let's first draw g(x), which does not have a hole:\r\n" );
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document.write( "In fact when x=-2 we have g(-2)=\"%28x%2B5%29%2F%28x-9%29\"=\"%28-2%2B5%29%2F%28-2-9%29\"=\"-3%2F11%29\"\r\n" );
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document.write( "But we don't want g(x), we want f(x), so we must put a hole in\r\n" );
document.write( "the curve at the point (-2,\"-3%2F11\"), for that is a removable\r\n" );
document.write( "discontinuity.  \r\n" );
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document.write( "Edwin
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