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

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

Answered by Tinkutara last updated on 26/Jul/17

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

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

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

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