document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1078578>Question 1078578</A>: <I><SPAN STYLE=\"word-break: break-all;\"> What is the slope for the line tangent to the circle x^2 + y^2 = 8 at the point (2, -2)? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #692999 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=692999\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> formula for circle at this point is y=-sqrt (8-x^2)<BR>\n" );
document.write( "The derivative is -2x/-2 sqrt(8-x^2); when x=2, derivative equals 2/sqrt(4)=1, so the slope is 1<BR>\n" );
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