From 1622c7daa22081b8b6a43633ad432daa3cd9113e Mon Sep 17 00:00:00 2001
From: Alexander Molnar <37451552+BTmathic@users.noreply.github.com>
Date: Thu, 21 Mar 2019 06:08:43 -0400
Subject: [PATCH] Fix errors (#34497)

Not all curves can be represented by functions (e.g., a circle) so they cannot simply be interchanged. Any line that "crosses" a graph touches it at one point but isn't tangent, and other minor typos/formatting addressed.
---
 .../equation-of-tangent-line/index.md         | 47 +++++++++++--------
 1 file changed, 28 insertions(+), 19 deletions(-)

diff --git a/guide/english/mathematics/equation-of-tangent-line/index.md b/guide/english/mathematics/equation-of-tangent-line/index.md
index 86538283e4..c8cfacaedf 100644
--- a/guide/english/mathematics/equation-of-tangent-line/index.md
+++ b/guide/english/mathematics/equation-of-tangent-line/index.md
@@ -3,34 +3,43 @@ title: Equation of Tangent Line
 ---
 ## Equation of Tangent Line
 
-A tangent line to a curve is a straight line that touches a curve, or a graph of a function, at only a single point. The tangent line represents the instantaneous rate of change of the function at that one point. The slope of the tangent line at a point on the function is equal to the derivative of the function at the same point.
+A tangent line to a function f(x) is a straight line that passes through the point (x<sub>0</sub>, f(x<sub>0</sub>)) and has slope f'(x<sub>0</sub>). The slope of the tangent line represents the instantaneous rate of change of the function at that point.
 
+### Finding the equation of a tangent line:
 
-### Finding Equation of the tangent line:
+To find the equation of a tangent line,
 
-To find the equation of tangent line to a curve at point x=x0, we need to find the following:
+<p align='center'>
+  y = mx + b,
+</p>
 
-1. Find the derivative of the function (i.e.derivative of the equation of curve).
-2. Find the value of the derivative by putting x=x0 , this will be the slope of the tangent (say m).
-3. Find the value y0, by putting the value of x0 in the equation of the curve. Our tangent will pass through this point (x0,y0)
-4. Find the equation of the tangent using point-slope form. As the tangent passes through (x0,y0) and have slope m, the equation of the tangent line can be given as:
-(y-y0)=m.(x-x0)
+of a function f(x) at point x = x<sub>0</sub>, we need to do the following:
 
-#### Example : To find the equation of tangent line to the curve f(x) = 4x^2-4x+1 at x=1
-Solution: 
-f(x) = 4x^2-4x+1
+1. Find the derivative of the function.
+2. Find the value of the derivative at x = x<sub>0</sub>, this will be the slope of the tangent (our m above).
+3. Find the value y<sub>0</sub> of the function at x<sub>0</sub>. The tangent will pass through the point (x<sub>0</sub>, y<sub>0</sub>).
+4. Find the equation of the tangent using point-slope form. As the tangent passes through (x<sub>0</sub>, y<sub>0</sub>) and has slope m, the equation of the tangent line can be written as (y - y<sub>0</sub>) = m(x - x<sub>0</sub>) or y = mx + (y<sub>0</sub> - mx<sub>0</sub>).
 
-Step 1 : f'(x) = 8x-4
+#### Example
 
-Step 2 : m = f'(2) = 8.2-4 = 12
+Find the equation of tangent line to the function f(x) = 4x<sup>2</sup> - 4x + 1 at x = 2.
 
-Step 3 : y0= f(x0) = f(2) = 4.2^2-4.2+1 = 16-8+1 = 9
+We proceed through the steps above.
 
-Step 4 : m=12 ;  (x0,y0)=(2,9)
+Step 1 : f'(x) = 8x - 4.
 
-Therefore, equation of tangent line is : 
-(y-y0)=m.(x-x0)
+Step 2 : m = f'(2) = 8 &middot; 2 - 4 = 12.
 
-=> (y-9)=12(x-2)
+Step 3 : y<sub>0</sub> = f(x<sub>0</sub>) = f(2) = 4 &middot; 2<sup>2</sup> - 4 &middot; 2 + 1 = 16 - 8 + 1 = 9.
 
-=> y=12x-15
+Step 4 : From steps 2 and 3 we have m = 12 and (x<sub>0</sub>, y<sub>0</sub>) = (2,9), so the equation of the tangent line is 
+
+<p alig='center'>
+  (y - 9) = 12(x - 2)
+</p>
+
+or, rearranging to slope intercept form,
+
+<p align='center'>
+  y = 12x - 15.
+</p>