document.write( "Question 1123335: How do you determine the point (x,y) at which the graph of \"+f%28x%29+=+%28-8x%29%2Fsqrt%282x-1%29+\" has a horizontal tangent?
\n" ); document.write( "So far I get d/dx to be \"%28-8%282x-1%29-8x%29%2F%28%282x-1%29%5E%283%2F2%29%29\" but what do you do to get a point value?
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Algebra.Com's Answer #739640 by josgarithmetic(39630)\"\" \"About 
You can put this solution on YOUR website!
Keep going.\r
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\n" ); document.write( "\n" ); document.write( "\"df%2Fdx=%28-8x%2B8%29%2F%28%282x-1%29sqrt%282x-1%29%29\"\r
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\n" ); document.write( "\n" ); document.write( "This derivative is 0 at \"x=1\", and \"f%281%29=-8\".
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