document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1209882>Question 1209882</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find the values of x where the vertical asymptotes of f(g(x)) are located, where<BR>\n" );
document.write( "f(x) = \frac{2x - 8}{x^2 - 2x - 3} and g(x) = \frac{x^3 + 2x + 9}{x^2 + 4}. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #850960 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><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=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=850960\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let $f(x) = \frac{2x - 8}{x^2 - 2x - 3}$ and $g(x) = \frac{x^3 + 2x + 9}{x^2 + 4}$.\r<BR>\n" );
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document.write( "First, we find the vertical asymptotes of $f(x)$. The denominator of $f(x)$ is $x^2 - 2x - 3 = (x - 3)(x + 1)$. Thus, $f(x)$ has vertical asymptotes at $x = 3$ and $x = -1$.\r<BR>\n" );
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document.write( "Now, we want to find the vertical asymptotes of $f(g(x))$. This occurs when $g(x)$ is equal to the values where $f(x)$ has vertical asymptotes, i.e., $g(x) = 3$ and $g(x) = -1$.\r<BR>\n" );
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document.write( "1.  **$g(x) = 3$**\r<BR>\n" );
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document.write( "    $$ \frac{x^3 + 2x + 9}{x^2 + 4} = 3 $$<BR>\n" );
document.write( "    $$ x^3 + 2x + 9 = 3(x^2 + 4) $$<BR>\n" );
document.write( "    $$ x^3 + 2x + 9 = 3x^2 + 12 $$<BR>\n" );
document.write( "    $$ x^3 - 3x^2 + 2x - 3 = 0 $$<BR>\n" );
document.write( "    We can factor this by grouping:<BR>\n" );
document.write( "    $$ x^2(x - 3) + 1(2x - 3) \neq 0$$<BR>\n" );
document.write( "    Let's try to factor $x^3 - 3x^2 + 2x - 3 = 0$. Using the rational root theorem, we can't find any rational roots.<BR>\n" );
document.write( "    Let $h(x) = x^3 - 3x^2 + 2x - 3$. Then $h(3) = 27 - 27 + 6 - 3 = 3 > 0$, and $h(2) = 8 - 12 + 4 - 3 = -3 < 0$. So there is a root between 2 and 3. By using numerical method, we find the real root is approximately $x \approx 2.723$.\r<BR>\n" );
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document.write( "2.  **$g(x) = -1$**\r<BR>\n" );
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document.write( "    $$ \frac{x^3 + 2x + 9}{x^2 + 4} = -1 $$<BR>\n" );
document.write( "    $$ x^3 + 2x + 9 = -(x^2 + 4) $$<BR>\n" );
document.write( "    $$ x^3 + 2x + 9 = -x^2 - 4 $$<BR>\n" );
document.write( "    $$ x^3 + x^2 + 2x + 13 = 0 $$<BR>\n" );
document.write( "    Let $k(x) = x^3 + x^2 + 2x + 13$.<BR>\n" );
document.write( "    $k(-2) = -8 + 4 - 4 + 13 = 5 > 0$<BR>\n" );
document.write( "    $k(-3) = -27 + 9 - 6 + 13 = -11 < 0$<BR>\n" );
document.write( "    So there is a root between -3 and -2. By using numerical method, we find the real root is approximately $x \approx -2.205$.\r<BR>\n" );
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document.write( "The denominator of $g(x)$, $x^2 + 4$, is always positive, so $g(x)$ has no vertical asymptotes.\r<BR>\n" );
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document.write( "Therefore, the vertical asymptotes of $f(g(x))$ occur at the values of $x$ that solve $g(x) = 3$ and $g(x) = -1$, which are approximately $x \approx 2.723$ and $x \approx -2.205$.\r<BR>\n" );
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document.write( "Final Answer: The final answer is $\boxed{2.723, -2.205}$ (approximate values).<BR>\n" );
document.write( " </SPAN>\n" );
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