# Ex 6.3,12

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

Ex 6.3,12 Find the equations of all lines having slope 0 which are tangent to the curve 𝑦 = 1𝑥2 −2𝑥 + 3 Given Curve is 𝑦 = 1𝑥2 −2𝑥 + 3 Slope of tangent is 𝑑𝑦𝑑𝑥 𝑑𝑦𝑑𝑥=𝑑1𝑥2 − 2𝑥 + 3𝑑𝑥 𝑑𝑦𝑑𝑥=𝑑𝑥2 − 2𝑥 + 3−1𝑑𝑥 𝑑𝑦𝑑𝑥=−1𝑥2−2𝑥+3−2 . 𝑑𝑥2− 2𝑥 + 3𝑑𝑥 𝑑𝑦𝑑𝑥=−𝑥2−2𝑥+3−22𝑥−2 𝑑𝑦𝑑𝑥=−2𝑥2−2𝑥+3−2𝑥−1 𝑑𝑦𝑑𝑥=−2𝑥 − 1𝑥2 − 2𝑥 + 32 Hence Slope of tangent is −2𝑥 − 1𝑥2 − 2𝑥 + 32 Given Slope of tangent is 0 ⇒ 𝑑𝑦𝑑𝑥=0 −2𝑥 − 1𝑥2 − 2𝑥 + 32=0 −2𝑥 − 1=0 ×𝑥2 − 2𝑥 + 32 −2𝑥 − 1=0 𝑥−1=0 𝑥=1 Finding y when 𝑥=1 𝑦=1𝑥2 − 2𝑥 + 3 𝑦=112 − 21 + 3 𝑦=11 − 2 + 3 𝑦=12 Point is 𝟏 , 𝟏𝟐 Thus , tangent passes through 1 , 12 Equation of tangent at 1 , 12 & having Slope zero is 𝑦 −12=0𝑥−1 𝑦 −12=0 𝑦=12 Hence Equation of tangent is 𝒚=𝟏𝟐

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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.