What is the tangent of y= 6/(x^2),(1,6) ?

The tangent of y= 6/(x^2),(1,6) is y=-12x+18

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tangent of y= 6/(x^2),(1,6)

tangent of

y = \frac{6}{x ^{2}}, (1, 6)

y = - 12 x + 18

Solution steps

Find the slope of

y = \frac{6}{x ^{2}} : \frac{d y}{d x} = - \frac{12}{x ^{3}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 12

Find the line with slope m=

- 12

and passing through

(1, 6) : y = - 12 x + 18

y = - 12 x + 18

\int \frac{1}{x ^{2} ( x ^{2} - 9 )} d x

y = 3 x^{3} - x^{2} + 5, (2, 25)

\frac{d}{d x} (2 x^{3} + 4 x^{2} - 3 x)

\int_{1}^{\infty} \frac{1}{x ^{4}} d x

\int e^{(x + 1)^{2}} (x + 1) d x

What is the tangent of y= 6/(x^2),(1,6) ?
The tangent of y= 6/(x^2),(1,6) is y=-12x+18