{"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

firstly, what is the slope of this line going to be? \u00a0anytime we are asked about slope, immediately find the derivative of the function<\/a><\/span>. \u00a0we should get y\u2019 = 3x<\/span>2<\/span><\/sup> \u2013 4x + 1. \u00a0evaluate this derivative at x = 1, and we get 3(<\/span>1<\/span>)2<\/sup> -4(1) +1 = 3-4+1= 0. \u00a0the slope, <\/span>m<\/span><\/i>, of this function at x=1 is <\/span>0<\/span><\/i>. \u00a0<\/span>m=0<\/span><\/i>. \u00a0(note, for the ap exam, you should also be able to use the derivative of this function in a similar way to find local minimums and maximums \u2013 we should be able to see that because our slope is 0, we are looking at a line that exists at a local minimum or maximum).\u00a0<\/span><\/p>\n

second, let us find a set of points (<\/span>x<\/span><\/i>1<\/span><\/i>, y<\/span><\/i>1<\/span><\/i>) <\/span><\/i>that exist on the line. \u00a0at this point, we can only use one value of x, and that is the value given, x=1. \u00a0to find the value y, we plug it into our original equation: y = (1)<\/span>3<\/span><\/sup>-2(1)<\/span>2<\/span><\/sup>+1-3 = 1-2+1-3 = -3. \u00a0therefore (<\/span>x<\/span><\/i>1<\/span><\/i>, y<\/span><\/i>1<\/span><\/i>) = (1, -3). \u00a0<\/span><\/i>we now have both a point on our line and the slope of our line.\u00a0 this is everything we need to find our equation.<\/span><\/p>\n

the equation of our line:<\/p>\n

y-y<\/span>1<\/span> = m(x-x<\/span>1<\/span>)<\/span><\/em><\/p>\n

y-(-3) = 0(x-1)<\/span><\/i><\/p>\n

y +3 = 0<\/span><\/i><\/p>\n

y = -3<\/i><\/b><\/p>\n

 <\/p>\n

here we have the equation with the tangent line drawn in:<\/span><\/p>\n

\"slope<\/p>\n

(can you find a local maximum of this function?)<\/p>\n

another tangent line equation example<\/h2>\n

let\u2019s do the exact same question as above, but at a new point:\u00a0<\/span>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=2.<\/b> \u00a0<\/span><\/p>\n

again, what is the slope of this line going to be? \u00a0first, the derivative: y\u2019 = 3x<\/span>2<\/span><\/sup> \u2013 4x + 1. \u00a0evaluate at x = 2.<\/span><\/p>\n

3(2)<\/span>2<\/span><\/sup>-4(2) +1 = 12-8+1 = 5. \u00a0the slope, <\/span>m<\/span><\/i>, of this function at x=2 is <\/span>5<\/i>\u00a0(<\/span>m=5)<\/span><\/i>. \u00a0<\/span><\/p>\n

 <\/p>\n

set of points (<\/span>x<\/span><\/i>1<\/span><\/i>, y<\/span><\/i>1<\/span><\/i>). \u00a0<\/span><\/i>we can only use x=2. \u00a0plug it into our original equation.<\/span><\/p>\n

y = \u00a02<\/span>3<\/span><\/sup>-2(2)<\/span>2<\/span><\/sup>+2-3 = 8-8+2-3 = -1.<\/span><\/p>\n

(<\/span>x<\/span><\/i>1<\/span><\/i>, y<\/span><\/i>1<\/span><\/i>) = (2, -1).<\/span><\/i><\/p>\n

 <\/p>\n

the equation of our line:<\/span><\/p>\n

y-y<\/span>1<\/span> = m(x-x<\/span>1<\/span>)<\/span><\/p>\n

y-(-1) = 5(x-2)<\/span><\/i><\/p>\n

y +1 = 5x-10<\/span><\/i><\/p>\n

y = 5x-11<\/i><\/b>
\n\"slope<\/p>\n

finding the slope of a tangent line: a review<\/span><\/h2>\n

finding the equation of a line tangent to a curve at a point always comes down to the following three steps:<\/span><\/p>\n

    \n
  1. find the derivative and use it to determine our slope m<\/i> at the point given<\/strong><\/li>\n
  2. determine the y\u00a0value of the function at the x value we are given.<\/strong><\/li>\n
  3. plug what we\u2019ve found into the equation of a line.<\/strong><\/li>\n<\/ol>\n

    master these steps, and we will be able to find the tangent line to any curve at any point.<\/span><\/p>\n","protected":false},"excerpt":{"rendered":"

    many questions on the ap calculus exam involve finding the equation of a tangent line. check out this post for a failsafe 3-step process for doing so!<\/p>\n","protected":false},"author":224,"featured_media":8277,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[240],"tags":[241,247,245,246],"ppma_author":[24930],"acf":[],"yoast_head":"\nhow to find the slope of a line tangent to a curve - magoosh blog | high school<\/title>\n<meta name=\"description\" content=\"many questions on the ap calculus exams involve finding the equation of a tangent line. check out this post for a failsafe 3-step process for doing so!\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"\/\/www.catharsisit.com\/hs\/ap\/slope-tangent-ap-calculus\/\" \/>\n<meta property=\"og:locale\" content=\"en_us\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"how to find the slope of a line tangent to a curve\" \/>\n<meta property=\"og:description\" content=\"many questions on the ap calculus exams involve finding the equation of a tangent line. check out this post for a failsafe 3-step process for doing so!\" \/>\n<meta property=\"og:url\" content=\"\/\/www.catharsisit.com\/hs\/ap\/slope-tangent-ap-calculus\/\" \/>\n<meta property=\"og:site_name\" content=\"magoosh blog | high school\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/magooshsat\/\" \/>\n<meta property=\"article:published_time\" content=\"2016-12-24t04:51:50+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2016-12-28t20:58:01+00:00\" \/>\n<meta property=\"og:image\" content=\"\/\/www.catharsisit.com\/hs\/files\/2016\/12\/screen-shot-2016-12-23-at-11.48.57-pm.png\" \/>\n\t<meta property=\"og:image:width\" content=\"327\" \/>\n\t<meta property=\"og:image:height\" content=\"330\" \/>\n\t<meta property=\"og:image:type\" content=\"image\/png\" \/>\n<meta name=\"author\" content=\"zachary\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@magooshsat_act\" \/>\n<meta name=\"twitter:site\" content=\"@magooshsat_act\" \/>\n<meta name=\"twitter:label1\" content=\"written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"zachary\" \/>\n\t<meta name=\"twitter:label2\" content=\"est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"3 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"article\",\"@id\":\"\/\/www.catharsisit.com\/hs\/ap\/slope-tangent-ap-calculus\/#article\",\"ispartof\":{\"@id\":\"\/\/www.catharsisit.com\/hs\/ap\/slope-tangent-ap-calculus\/\"},\"author\":{\"name\":\"zachary\",\"@id\":\"\/\/www.catharsisit.com\/hs\/#\/schema\/person\/2bb580c8b05f06690589778ec3805636\"},\"headline\":\"how to find the slope of a line tangent to a curve\",\"datepublished\":\"2016-12-24t04:51:50+00:00\",\"datemodified\":\"2016-12-28t20:58:01+00:00\",\"mainentityofpage\":{\"@id\":\"\/\/www.catharsisit.com\/hs\/ap\/slope-tangent-ap-calculus\/\"},\"wordcount\":612,\"commentcount\":2,\"publisher\":{\"@id\":\"\/\/www.catharsisit.com\/hs\/#organization\"},\"keywords\":[\"ap calculus\",\"derivatives\",\"equation of a line\",\"tangent to curve\"],\"articlesection\":[\"ap\"],\"inlanguage\":\"en-us\",\"potentialaction\":[{\"@type\":\"commentaction\",\"name\":\"comment\",\"target\":[\"\/\/www.catharsisit.com\/hs\/ap\/slope-tangent-ap-calculus\/#respond\"]}]},{\"@type\":\"webpage\",\"@id\":\"\/\/www.catharsisit.com\/hs\/ap\/slope-tangent-ap-calculus\/\",\"url\":\"\/\/www.catharsisit.com\/hs\/ap\/slope-tangent-ap-calculus\/\",\"name\":\"how to find the slope of a line tangent to a curve - 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