Example 18
Last updated at Dec. 8, 2016 by Teachoo
Transcript
Example 18 Find the equation of the tangent to the curve y = 𝑥 − 7𝑥 − 2(𝑥 − 3) at the point where it cuts the x-axis. Slope of the tangent to the curve is 𝑑𝑦𝑑𝑥 = 𝑥 − 7′ 𝑥 − 2 𝑥 − 3 − 𝑥 − 7 𝑥 − 3 + 𝑥 − 2′𝑥 − 22 𝑥 − 32 𝑑𝑦𝑑𝑥 = 1 𝑥 − 2 𝑥 − 3 − 𝑥 − 71𝑥 − 3 + 𝑥 − 2𝑥 − 22 𝑥 − 32 𝑑𝑦𝑑𝑥 = 𝑥 − 2 𝑥 − 3 − 𝑥 − 72𝑥 − 5𝑥 − 22 𝑥 − 32 𝑑𝑦𝑑𝑥 = 1 − 𝑦2𝑥 − 5𝑥 − 2𝑥 − 3 At the point where the curve cuts the x axis, y = 0 Hence the equation of the curve is 0 = 𝑥 − 7𝑥 − 2𝑥 − 3 ⇒ x = 7 The curve cuts the x – axis at point (7, 0) At point (7, 0) the slope is 𝑑𝑦𝑑𝑥7, 0 = 1 − 027−57 − 2 7 − 3 = 15 × 4 = 120 Now, the equation of the tangent at point (7, 0) with slope 120 is 𝑦 −𝑦1= 𝑚 𝑥 − 𝑥1 𝑦−0= 120 𝑥−7 20𝑦=𝑥−7 𝟐𝟎𝒚−𝒙+𝟕=𝟎
Examples
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Example 18 You are here
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About the Author
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.