{"id":8275,"date":"2016-12-23t20:51:50","date_gmt":"2016-12-24t04:51:50","guid":{"rendered":"\/\/www.catharsisit.com\/hs\/?p=8275"},"modified":"2016-12-28t12:58:01","modified_gmt":"2016-12-28t20:58:01","slug":"slope-tangent-ap-calculus","status":"publish","type":"post","link":"\/\/www.catharsisit.com\/hs\/ap\/slope-tangent-ap-calculus\/","title":{"rendered":"how to find the slope of a line tangent to a curve"},"content":{"rendered":"
many common questions asked on the ap calculus exams involve finding the equation of a line tangent to a curve at a point. \u00a0if we are adept at quickly taking derivatives of functions, then 90 percent of the work for these types of problems is done. \u00a0everything else comes down to quick algebra.<\/p>\n
the first thing we need to do is to go back to what we learned in our algebra: the equation of a line or y = mx+b<\/strong>, where <\/span>m<\/span><\/i> is our slope and <\/span>b<\/span><\/i> is our y-intercept. \u00a0this should be at our fingertips. \u00a0now, we don\u2019t always have our y-intercept, so a slightly different form of our equation of a line is often useful: y-y<\/span>1<\/span> = m(x-x<\/span>1<\/span>), where <\/span>m<\/span><\/i> is our slope, and <\/span>x<\/span><\/i>1<\/span><\/i> and <\/span>y<\/span><\/i>1<\/span><\/i> are the coordinates of a point. \u00a0questions involving finding the equation of a line tangent to a point then come down to two parts: finding the slope, and finding a point on the line.<\/span><\/p>\n
let us take an example<\/span><\/h2>\n
find the equations of a line tangent to y = x<\/b>3<\/b><\/sup>-2x<\/b>2<\/b><\/sup>+x-3 at the point x=1.<\/b> \u00a0<\/span><\/p>\n