Question-108104

Question Number 108104 by bemath last updated on 14/Aug/20

Answered by bobhans last updated on 14/Aug/20

\frac{B ob H ans}{★}

(1) let the curve y = f (x) with gradient = \frac{dy}{dx}

where \frac{dy}{dx} = k (\frac{y - 2}{x - 1}) or \frac{dy}{y - 2} = k \frac{dx}{x - 1}

\int \frac{dy}{y - 2} = \int \frac{k dx}{x - 1} \Rightarrow \ln (y - 2) = k \ln (x - 1) + c

substitute the point (2, 5) & (9, 8)

\Rightarrow {\begin{cases} (2, 5) \to \ln (3) = k . \ln (1) + c, c = \ln (3) \\ (9, 8) \to \ln (6) = k . \ln (8) + \ln (3) \Rightarrow k = \frac{1}{3} \end{cases}

therefore \Rightarrow \ln (y - 2) = \frac{1}{3} \ln (x - 1) + \ln (3)

y - 2 = 3 . \sqrt[3]{x - 1} .

Commented by bemath last updated on 14/Aug/20

t q m r b o b

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