document.write( "Question 383320: I am so confused with this. Please help.\r
\n" ); document.write( "\n" ); document.write( "Graph y=log(base 3)(x+2)+1
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Algebra.Com's Answer #271441 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
Graph y=log3(x+2)+1\r
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\n" ); document.write( "Identify the Domain:
\n" ); document.write( "Solve x+2 > 0
\n" ); document.write( "x > -2 is the domain
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\n" ); document.write( "y = [log3(x+2)] + 1
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\n" ); document.write( "Vertical asymptote at x = -2
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\n" ); document.write( "x-intercept:
\n" ); document.write( "Let y = 0
\n" ); document.write( "Then [log3(x+2)] + 1 = 0
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\n" ); document.write( "[log3(x+2)] = -1
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\n" ); document.write( "x+2 = 3^-1
\n" ); document.write( "x+2 = 1/3
\n" ); document.write( "x = -5/3
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\n" ); document.write( "Plot the asymptote and the point (-5/3,0)
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\n" ); document.write( "Sketch the log curve using those two reference facts.
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.\r
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