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.
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The Coding Aviator
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@ -3,34 +3,43 @@ title: Equation of Tangent Line
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## Equation of Tangent Line
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## Equation of Tangent Line
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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.
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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.
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### Finding the equation of a tangent line:
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### Finding Equation of the tangent line:
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To find the equation of a tangent line,
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To find the equation of tangent line to a curve at point x=x0, we need to find the following:
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<p align='center'>
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y = mx + b,
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</p>
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1. Find the derivative of the function (i.e.derivative of the equation of curve).
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of a function f(x) at point x = x<sub>0</sub>, we need to do the following:
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2. Find the value of the derivative by putting x=x0 , this will be the slope of the tangent (say m).
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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)
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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:
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(y-y0)=m.(x-x0)
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#### Example : To find the equation of tangent line to the curve f(x) = 4x^2-4x+1 at x=1
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1. Find the derivative of the function.
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Solution:
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2. Find the value of the derivative at x = x<sub>0</sub>, this will be the slope of the tangent (our m above).
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f(x) = 4x^2-4x+1
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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>).
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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>).
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Step 1 : f'(x) = 8x-4
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#### Example
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Step 2 : m = f'(2) = 8.2-4 = 12
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Find the equation of tangent line to the function f(x) = 4x<sup>2</sup> - 4x + 1 at x = 2.
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Step 3 : y0= f(x0) = f(2) = 4.2^2-4.2+1 = 16-8+1 = 9
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We proceed through the steps above.
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Step 4 : m=12 ; (x0,y0)=(2,9)
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Step 1 : f'(x) = 8x - 4.
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Therefore, equation of tangent line is :
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Step 2 : m = f'(2) = 8 · 2 - 4 = 12.
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(y-y0)=m.(x-x0)
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=> (y-9)=12(x-2)
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Step 3 : y<sub>0</sub> = f(x<sub>0</sub>) = f(2) = 4 · 2<sup>2</sup> - 4 · 2 + 1 = 16 - 8 + 1 = 9.
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=> y=12x-15
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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
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<p alig='center'>
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(y - 9) = 12(x - 2)
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</p>
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or, rearranging to slope intercept form,
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<p align='center'>
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y = 12x - 15.
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</p>
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