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Fix errors (#34497)

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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.
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Alexander Molnar
2019-03-21 06:08:43 -04:00
committed by The Coding Aviator
parent 419a7151f2
commit 1622c7daa2
1 changed files with 28 additions and 19 deletions
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guide/english/mathematics/equation-of-tangent-line/index.md
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@ -3,34 +3,43 @@ title: Equation of Tangent Line
--- ---
## 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). of a function f(x) at point x = x<sub>0</sub>, we need to do the following:
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)
#### Example : To find the equation of tangent line to the curve f(x) = 4x^2-4x+1 at x=1 1. Find the derivative of the function.
Solution: 2. Find the value of the derivative at x = x<sub>0</sub>, this will be the slope of the tangent (our m above).
f(x) = 4x^2-4x+1 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 : Step 2 : m = f'(2) = 8 &middot; 2 - 4 = 12.
(y-y0)=m.(x-x0)
=> (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>