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Question Number 147753 by Odhiambojr last updated on 23/Jul/21

Find a point on the curve   y=x^2 −2x−3 at which the tangent is   parallel to the x-axis

$${Find}\:{a}\:{point}\:{on}\:{the}\:{curve} \\ $$$$\:{y}={x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{3}\:{at}\:{which}\:{the}\:{tangent}\:{is} \\ $$$$\:{parallel}\:{to}\:{the}\:{x}-{axis} \\ $$

Answered by Olaf_Thorendsen last updated on 23/Jul/21

(dy/dx) = 2x−2 = 2(x−1)  (dy/dx) = 0  ⇔ x = 1 and then y = −4

$$\frac{{dy}}{{dx}}\:=\:\mathrm{2}{x}−\mathrm{2}\:=\:\mathrm{2}\left({x}−\mathrm{1}\right) \\ $$$$\frac{{dy}}{{dx}}\:=\:\mathrm{0}\:\:\Leftrightarrow\:{x}\:=\:\mathrm{1}\:\mathrm{and}\:\mathrm{then}\:{y}\:=\:−\mathrm{4} \\ $$

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