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Example 18 - Find equation of tangent at point where it cuts - Examples

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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise
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Example 18 Find the equation of the tangent to the curve y = ﷐𝑥 − 7﷮﷐𝑥 − 2﷯(𝑥 − 3)﷯ at the point where it cuts the x-axis. Slope of the tangent to the curve is ﷐𝑑𝑦﷮𝑑𝑥﷯ = ﷐﷐﷐𝑥 − 7﷯﷮′﷯ ﷐𝑥 − 2﷯ ﷐𝑥 − 3﷯ − ﷐𝑥 − 7﷯ ﷐﷐﷐𝑥 − 3﷯ + ﷐𝑥 − 2﷯﷯﷮′﷯﷮﷐﷐𝑥 − 2﷯﷮2﷯ ﷐﷐𝑥 − 3﷯﷮2﷯﷯ ﷐𝑑𝑦﷮𝑑𝑥﷯ = ﷐﷐1﷯ ﷐𝑥 − 2﷯ ﷐𝑥 − 3﷯ − ﷐𝑥 − 7﷯﷐﷐1﷯﷐𝑥 − 3﷯ + ﷐𝑥 − 2﷯﷯﷮﷐﷐𝑥 − 2﷯﷮2﷯ ﷐﷐𝑥 − 3﷯﷮2﷯﷯ ﷐𝑑𝑦﷮𝑑𝑥﷯ = ﷐﷐𝑥 − 2﷯ ﷐𝑥 − 3﷯ − ﷐𝑥 − 7﷯﷐2𝑥 − 5﷯﷮﷐﷐𝑥 − 2﷯﷮2﷯ ﷐﷐𝑥 − 3﷯﷮2﷯﷯ ﷐𝑑𝑦﷮𝑑𝑥﷯ = ﷐1 − 𝑦﷐2𝑥 − 5﷯﷮﷐𝑥 − 2﷯﷐𝑥 − 3﷯﷯ At the point where the curve cuts the x axis, y = 0 Hence the equation of the curve is 0 = ﷐𝑥 − 7﷮﷐𝑥 − 2﷯﷐𝑥 − 3﷯﷯ ⇒ x = 7 The curve cuts the x – axis at point (7, 0) At point (7, 0) the slope is ﷐﷐﷐𝑑𝑦﷮𝑑𝑥﷯﷯﷮﷐7, 0﷯﷯ = ﷐1 − 0﷐2﷐7﷯−5﷯﷮﷐7 − 2﷯ ﷐7 − 3﷯﷯ = ﷐1﷮5 × 4﷯ = ﷐1﷮20﷯ Now, the equation of the tangent at point (7, 0) with slope ﷐1﷮20﷯ is 𝑦 −﷐𝑦﷮1﷯= 𝑚 ﷐𝑥 − 𝑥1﷯ 𝑦−0= ﷐1﷮20﷯ ﷐𝑥−7﷯ 20𝑦=𝑥−7 𝟐𝟎𝒚−𝒙+𝟕=𝟎

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