document.write( "Question 1210370: Using the discriminant method, find the range of the function:\r
\n" ); document.write( "\n" ); document.write( "y =(x+1)/(sqrt(x² - 5)) \n" ); document.write( "
Algebra.Com's Answer #852108 by mccravyedwin(407)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "I corrected my former answer.\r\n" );
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document.write( "\"y=%28x%2B1%5E%22%22%29%2Fsqrt%28x%5E2-5%29\"\r\n" );
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document.write( "There are vertical asymptotes at \"x=%22%22+%2B-+sqrt%285%29\",\r\n" );
document.write( "and there are no points between those two vertical asymptotes.\r\n" );
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document.write( "We know that there is a vertical asymptote at \"x=sqrt%285%29\" so y goes to infinity on the right.  \r\n" );
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document.write( "Also since the limit of y as x approaches +infinity is 1, there is a horizontal\r\n" );
document.write( "asymptote y = 1 and y is always above that horizontal asymptote when\r\n" );
document.write( "\"x%3Esqrt%285%29\"\r\n" );
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document.write( "Also since the limit of y as x approaches -infinity is -1, there is a horizontal\r\n" );
document.write( "asymptote y = -1 and y is always above that horizontal asymptote when\r\n" );
document.write( "\"x%3Csqrt%285%29\"\r\n" );
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document.write( "\"y%2Asqrt%28x%5E2-5%29=x%2B1\"\r\n" );
document.write( "\"y%5E2%28x%5E2-5%29=%28x%2B1%29%5E2\"\r\n" );
document.write( "\"y%5E2x%5E2-5y%5E2+=+x%5E2%2B2x%2B1\"\r\n" );
document.write( "\"y%5E2x%5E2+-+x%5E2-+5y%5E2-2x-1=0\"\r\n" );
document.write( "\"x%5E2%28y%5E2-1%29-2x-%285y%5E2%2B1%29+=+0\"\r\n" );
document.write( "\"Discriminant+=+B%5E2-4AC+=+%28-2%29%5E2-4%28y%5E2-1%29%28-%285y%5E2%2B1%29%5E%22%22%29\"\r\n" );
document.write( "\"Discriminant+=+4-4%28y%5E2-1%29%28-5y%5E2-1%29\"\r\n" );
document.write( "Factor the negative out of the last factor:\r\n" );
document.write( "\"Discriminant+=+4%2B4%28y%5E2-1%29%285y%5E2%2B1%29\"\r\n" );
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document.write( "\"Discriminant+=+20y%5E4-16y%5E2\"\r\n" );
document.write( "The discriminant is not negative:\r\n" );
document.write( "\"20y%5E4-16y%5E2%3E=0\"\r\n" );
document.write( "\"4y%5E2%285y%5E2-4%29%3E=0\"\r\n" );
document.write( "\"y%5E2%285y%5E2-4%29%3E=0\"\r\n" );
document.write( "Since \"y%5E2%3E=0\" is always true, we must have\r\n" );
document.write( "\"5y%5E2-4%3E=0\"\r\n" );
document.write( "\"5y%5E2%3E=4\"\r\n" );
document.write( "\"y%5E2%3E=4%2F5\"\r\n" );
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document.write( "Since y is always negative left of its left asymptote\r\n" );
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document.write( "\"y%3C-sqrt%284%2F5%29\"\r\n" );
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document.write( "However y is always above its horizontal asymptote y = 1 to the right of its\r\n" );
document.write( "vertical asymptote \r\n" );
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document.write( "Since the denominator is always positive, y is always negative when x<-1 \r\n" );
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document.write( "So on the left of the vertical asympote. \"y%3C=-sqrt%284%2F5%29\" \r\n" );
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document.write( "However on the right of its horizontal asymptote \"x=sqrt%285%29\" y is always\r\n" );
document.write( "above its horizontal asymptote y = 1, so y > 1 when \"x%3Esqrt%285%29\"\r\n" );
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document.write( "So the range of the function is\r\n" );
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document.write( "or you can rationalize the denominator:\r\n" );
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document.write( "Edwin
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