{"id":106,"date":"2006-01-26T21:26:37","date_gmt":"2006-01-26T21:26:37","guid":{"rendered":"http:\/\/www.sixthform.info\/maths\/?p=106"},"modified":"2011-05-12T09:46:34","modified_gmt":"2011-05-12T08:46:34","slug":"cubics","status":"publish","type":"post","link":"https:\/\/www.sixthform.info\/maths\/?p=106","title":{"rendered":"Cubics"},"content":{"rendered":"<p>A problem that I give some of the students is to show that for a cubic <img src='\/maths\/latexrender\/pictures\/8fa14cdd754f91cc6554c9e71929cce7.gif' title='f' alt='f' align=absmiddle> that if <img src='\/maths\/latexrender\/pictures\/3e7ad5435b872d20e4987e4f8117bad6.gif' title='x=\\alpha' alt='x=\\alpha' align=absmiddle> and <img src='\/maths\/latexrender\/pictures\/7eefdf5737e77d20bf30280aa853dcd0.gif' title='x=\\beta' alt='x=\\beta' align=absmiddle> are the x-coordinates of its stationary points then <img src='\/maths\/latexrender\/pictures\/f533cfc7cac17e9154b7e232fcff708b.gif' title='f^{\\prime\\prime}( \\alpha ) =-f^{\\prime \\prime}( \\beta )' alt='f^{\\prime\\prime}( \\alpha ) =-f^{\\prime \\prime}( \\beta )' align=absmiddle>. Hopefully, they will have already noticed this result in the exercises they have done.<\/p>\n<p>I then ask them to show that the point of inflexion of <img src='\/maths\/latexrender\/pictures\/8fa14cdd754f91cc6554c9e71929cce7.gif' title='f' alt='f' align=absmiddle> is at the midpoint of the turning points. This involves using the equations for the roots of a quadratic: <img src='\/maths\/latexrender\/pictures\/75217c1633426504214d83b7ba90b0b7.gif' title='\\alpha + \\beta=-\\frac{b}{a}' alt='\\alpha + \\beta=-\\frac{b}{a}' align=absmiddle> and <img src='\/maths\/latexrender\/pictures\/2ee9012b26089071db79ca8dc7f10661.gif' title='\\alpha \\beta=\\frac{c}{a}' alt='\\alpha \\beta=\\frac{c}{a}' align=absmiddle> but the algebra isn&#8217;t particularly nice (details are <a href=\"http:\/\/sixthform.info\/maths\/files\/harder_questions.pdf\" target=\"_blank\">here<\/a>).<\/p>\n<p>It dawned on me that there&#8217;s a better way which is not only simpler to prove but gives a more powerful result: <em>a cubic is (rotationally) symmetric about its point of inflexion<\/em>.<\/p>\n<p>To see this, take the cubic and translate it so that the point of inflexion is at the origin. This clearly has no effect on the symmetry so it is sufficient to prove the result in this case.<\/p>\n<p>If <img src='\/maths\/latexrender\/pictures\/5d59a60b40cfedc929a43400c07f8c6b.gif' title='f(x)=ax^3+bx^2+cx+d' alt='f(x)=ax^3+bx^2+cx+d' align=absmiddle> has its point of inflexion at the origin then clearly <img src='\/maths\/latexrender\/pictures\/13285769c8cd293bfd541406e0621d87.gif' title='d=0' alt='d=0' align=absmiddle> and, as <img src='\/maths\/latexrender\/pictures\/788d166efc1e38776d740b3f0f6b5f1f.gif' title='f^{\\prime\\prime}(0)=2b' alt='f^{\\prime\\prime}(0)=2b' align=absmiddle>, it follows that <img src='\/maths\/latexrender\/pictures\/f6d5eef5ee5e51fc839bb54201c62e3b.gif' title='b=0' alt='b=0' align=absmiddle> so <img src='\/maths\/latexrender\/pictures\/162d2b1e3c65ee219f2c3660b6f7e488.gif' title='f(x)=ax^3+cx' alt='f(x)=ax^3+cx' align=absmiddle>. Rotate this by <img src='\/maths\/latexrender\/pictures\/4f08e3dba63dc6d40b22952c7a9dac6d.gif' title='\\pi' alt='\\pi' align=absmiddle> about the origin, for example by applying the usual matrix <img src='\/maths\/latexrender\/pictures\/5d3dbaac9c58f893c70d512189007343.gif' title='\\begin{pmatrix} -1 &amp; 0 \\\\ 0 &amp; -1 \\end{pmatrix}' alt='\\begin{pmatrix} -1 &amp; 0 \\\\ 0 &amp; -1 \\end{pmatrix}' align=absmiddle>, and you get <img src='\/maths\/latexrender\/pictures\/6100dbb3f9f33055d3c49d1504da48cd.gif' title='-f(x)=-ax^3-cx' alt='-f(x)=-ax^3-cx' align=absmiddle> which is clearly the same cubic. Hence it has rotational symmetry about the origin, and in particular, the point of inflexion at the origin is the midpoint of the stationary points.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>A problem that I give some of the students is to show that for a cubic that if and are the x-coordinates of its stationary points then . Hopefully, they will have already noticed this result in the exercises they have done. I then ask them to show that the point of inflexion of is [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-106","post","type-post","status-publish","format-standard","hentry","category-articles"],"_links":{"self":[{"href":"https:\/\/www.sixthform.info\/maths\/index.php?rest_route=\/wp\/v2\/posts\/106","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.sixthform.info\/maths\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.sixthform.info\/maths\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.sixthform.info\/maths\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.sixthform.info\/maths\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=106"}],"version-history":[{"count":2,"href":"https:\/\/www.sixthform.info\/maths\/index.php?rest_route=\/wp\/v2\/posts\/106\/revisions"}],"predecessor-version":[{"id":192,"href":"https:\/\/www.sixthform.info\/maths\/index.php?rest_route=\/wp\/v2\/posts\/106\/revisions\/192"}],"wp:attachment":[{"href":"https:\/\/www.sixthform.info\/maths\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=106"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.sixthform.info\/maths\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=106"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.sixthform.info\/maths\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=106"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}