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Question Number 18642 by tawa tawa last updated on 26/Jul/17

What is the area of the region bounded by coordinate axis and the line  tangent to the graph,  y = (1/8)x^2  + (1/2)x + 1, at the point  (0, 1)

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{region}\:\mathrm{bounded}\:\mathrm{by}\:\mathrm{coordinate}\:\mathrm{axis}\:\mathrm{and}\:\mathrm{the}\:\mathrm{line} \\ $$$$\mathrm{tangent}\:\mathrm{to}\:\mathrm{the}\:\mathrm{graph},\:\:\mathrm{y}\:=\:\frac{\mathrm{1}}{\mathrm{8}}\mathrm{x}^{\mathrm{2}} \:+\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}\:+\:\mathrm{1},\:\mathrm{at}\:\mathrm{the}\:\mathrm{point}\:\:\left(\mathrm{0},\:\mathrm{1}\right) \\ $$

Answered by Tinkutara last updated on 26/Jul/17

(dy/dx) = (x/4) + (1/2)  Slope of the tangent at x = 0 is (1/2).  ∴ Equation of tangent is  y − 1 = (x/2) ⇒ y = (x/2) + 1  Intercepts of this line are (0, 1) and  (−2, 0).  ∴ Area of this triangle = (1/2)×2×1  = 1 sq. unit

$$\frac{{dy}}{{dx}}\:=\:\frac{{x}}{\mathrm{4}}\:+\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{Slope}\:\mathrm{of}\:\mathrm{the}\:\mathrm{tangent}\:\mathrm{at}\:{x}\:=\:\mathrm{0}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{2}}. \\ $$$$\therefore\:\mathrm{Equation}\:\mathrm{of}\:\mathrm{tangent}\:\mathrm{is} \\ $$$${y}\:−\:\mathrm{1}\:=\:\frac{{x}}{\mathrm{2}}\:\Rightarrow\:{y}\:=\:\frac{{x}}{\mathrm{2}}\:+\:\mathrm{1} \\ $$$$\mathrm{Intercepts}\:\mathrm{of}\:\mathrm{this}\:\mathrm{line}\:\mathrm{are}\:\left(\mathrm{0},\:\mathrm{1}\right)\:\mathrm{and} \\ $$$$\left(−\mathrm{2},\:\mathrm{0}\right). \\ $$$$\therefore\:\mathrm{Area}\:\mathrm{of}\:\mathrm{this}\:\mathrm{triangle}\:=\:\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}×\mathrm{1} \\ $$$$=\:\mathrm{1}\:\mathrm{sq}.\:\mathrm{unit} \\ $$

Commented by tawa tawa last updated on 26/Jul/17

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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