{"id":8275,"date":"2016-12-23T20:51:50","date_gmt":"2016-12-24T04:51:50","guid":{"rendered":"https:\/\/magoosh.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":"https:\/\/magoosh.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],"class_list":["post-8275","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-ap","tag-ap-calculus","tag-derivatives","tag-equation-of-a-line","tag-tangent-to-curve"],"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. 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