document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=135069>Question 135069</A>: <I><SPAN STYLE=\"word-break: break-all;\"> This is not a text book question\r<BR>\n" );
document.write( "\n" );
document.write( "Given F(x) = 4x^2-1/x^2 -16<BR>\n" );
document.write( "a) find x intercepts<BR>\n" );
document.write( "b) y intercepts<BR>\n" );
document.write( "c) Find the vertical asymptote (write as equation of a line)<BR>\n" );
document.write( "d) Find the horizontal asymptote (write as equation of line)<BR>\n" );
document.write( "e) For x>4, is there a value which F(x) cannot exceed, and/or value which F(x) cannot fall below? Explain<BR>\n" );
document.write( "f) Find maximum or minimum in interval and state whether it is a maximum or minimum value for that interval<BR>\n" );
document.write( "g) Describe what happens on graph when x approaches infinity.\r<BR>\n" );
document.write( "\n" );
document.write( "I have several questions like this, but need direction on one, to help with the rest. I have the vertical asymptotes as 4 and -4 and the horizontal as 4.<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #98918 by </B><A HREF=/tutors/aboutme.mpl?userid=solver91311><B>solver91311(24713)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solver91311><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=solver91311><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=98918\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=F%28x%29+=+%284x%5E2-1%29%2F%28x%5E2+-16%29 ALIGN=MIDDLE  ALT=\"F%28x%29+=+%284x%5E2-1%29%2F%28x%5E2+-16%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "x-intercepts are where F(x)=0.  Since we know that <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=a%2Fb=0 ALIGN=MIDDLE  ALT=\"a%2Fb=0\"   DISABLED_style= \"max-width:90%; height:auto;\" > if and only if <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=a=0 ALIGN=MIDDLE  ALT=\"a=0\"   DISABLED_style= \"max-width:90%; height:auto;\" > AND <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=b%3C%3E0 ALIGN=MIDDLE  ALT=\"b%3C%3E0\"   DISABLED_style= \"max-width:90%; height:auto;\" >, all we need to to is set the numerator equal to zero, solve, and exclude any root that would make the denominator go to zero as well.  The exclusion bit won't be a problem here, but you always need to consider the situation.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=4x%5E2-1=0 ALIGN=MIDDLE  ALT=\"4x%5E2-1=0\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E2=1%2F4 ALIGN=MIDDLE  ALT=\"x%5E2=1%2F4\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x=1%2F2 ALIGN=MIDDLE  ALT=\"x=1%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" > or <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x=-1%2F2 ALIGN=MIDDLE  ALT=\"x=-1%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  No exclusions required, so the intercepts are (<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1%2F2 ALIGN=MIDDLE  ALT=\"1%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" >,<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=0 ALIGN=MIDDLE  ALT=\"0\"   DISABLED_style= \"max-width:90%; height:auto;\" >) and (<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-1%2F2 ALIGN=MIDDLE  ALT=\"-1%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" >,<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=0 ALIGN=MIDDLE  ALT=\"0\"   DISABLED_style= \"max-width:90%; height:auto;\" >).\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "The y-intercept is at F(0), so evaluate <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=F%280%29=%284%280%29%5E2-1%29%2F%280%5E2-16%29=%28-1%29%2F%28-16%29=1%2F16 ALIGN=MIDDLE  ALT=\"F%280%29=%284%280%29%5E2-1%29%2F%280%5E2-16%29=%28-1%29%2F%28-16%29=1%2F16\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  Hence the y-intercept is (<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=0 ALIGN=MIDDLE  ALT=\"0\"   DISABLED_style= \"max-width:90%; height:auto;\" >,<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1%2F16 ALIGN=MIDDLE  ALT=\"1%2F16\"   DISABLED_style= \"max-width:90%; height:auto;\" >)\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "You have the right numbers for the asymptotes, but you are instructed to express them as equations, so you want <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x=4 ALIGN=MIDDLE  ALT=\"x=4\"   DISABLED_style= \"max-width:90%; height:auto;\" >, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x=-4 ALIGN=MIDDLE  ALT=\"x=-4\"   DISABLED_style= \"max-width:90%; height:auto;\" > for the verticals, and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y+=4 ALIGN=MIDDLE  ALT=\"y+=4\"   DISABLED_style= \"max-width:90%; height:auto;\" > for the horiziontal.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "When <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%3E4 ALIGN=MIDDLE  ALT=\"x%3E4\"   DISABLED_style= \"max-width:90%; height:auto;\" > (or <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%3C-4 ALIGN=MIDDLE  ALT=\"x%3C-4\"   DISABLED_style= \"max-width:90%; height:auto;\" > for that matter), F will always be a little more than 4.  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=4x%5E2%3E1 ALIGN=MIDDLE  ALT=\"4x%5E2%3E1\"   DISABLED_style= \"max-width:90%; height:auto;\" > when <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%3E4 ALIGN=MIDDLE  ALT=\"x%3E4\"   DISABLED_style= \"max-width:90%; height:auto;\" >, so the numerator is positive, and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E2%3E16 ALIGN=MIDDLE  ALT=\"x%5E2%3E16\"   DISABLED_style= \"max-width:90%; height:auto;\" > when <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%3E4 ALIGN=MIDDLE  ALT=\"x%3E4\"   DISABLED_style= \"max-width:90%; height:auto;\" >, so the denominator is also positive.  F(5) = 11 > 4 (you can calculate this out if you like) meaning that F > 4 when <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%3E4 ALIGN=MIDDLE  ALT=\"x%3E4\"   DISABLED_style= \"max-width:90%; height:auto;\" > near <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x=4 ALIGN=MIDDLE  ALT=\"x=4\"   DISABLED_style= \"max-width:90%; height:auto;\" > and the constant terms will have less and less impact on the value of F as x gets larger.  Further, x would have to be infinite in order for F to equal 4 if <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%3E4 ALIGN=MIDDLE  ALT=\"x%3E4\"   DISABLED_style= \"max-width:90%; height:auto;\" > or <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%3C-4 ALIGN=MIDDLE  ALT=\"x%3C-4\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  That pretty much answers parts e and g.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=drawing%28600%2C600%2C-10%2C10%2C-10%2C10%2C%0D%0Agrid%281%29%2C%0D%0Agraph%28600%2C600%2C-10%2C10%2C-10%2C10%2C%284x%5E2-1%29%2F%28x%5E2+-16%29%29%0D%0A%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Ignore the vertical (sort of) lines that intersect the x-axis at about -4 and 4.  These are extraneous -- a shortcoming of the graph rendering system on this site.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Notice that the function has a local maximum on the interval (-4,4).  Fortunately it happens to coincide with the y-intercept, i.e. the local maximum is at (<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=0 ALIGN=MIDDLE  ALT=\"0\"   DISABLED_style= \"max-width:90%; height:auto;\" >,<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1%2F16 ALIGN=MIDDLE  ALT=\"1%2F16\"   DISABLED_style= \"max-width:90%; height:auto;\" >).\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "If you actually need to prove that the local extrema is at (<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=0 ALIGN=MIDDLE  ALT=\"0\"   DISABLED_style= \"max-width:90%; height:auto;\" >,<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1%2F16 ALIGN=MIDDLE  ALT=\"1%2F16\"   DISABLED_style= \"max-width:90%; height:auto;\" >), you will have to take the first derivative of the function and set it equal to zero to find the point.  Trying to render the quotient rule on this system is a horror, so I won't do it here.  Write back if you want to see it.\r<BR>\n" );
document.write( "\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );