What is the tangent of y= 9/(x^2+2),(1,3) ?

The tangent of y= 9/(x^2+2),(1,3) is y=-2x+5

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tangent of y= 9/(x^2+2),(1,3)

tangent of

y = \frac{9}{x ^{2} + 2}, (1, 3)

y = - 2 x + 5

Solution steps

Find the slope of

y = \frac{9}{x ^{2} + 2} : \frac{d y}{d x} = - \frac{18 x}{( x ^{2} + 2 ) ^{2}}

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 = - 2

Find the line with slope m=

- 2

and passing through

(1, 3) : y = - 2 x + 5

y = - 2 x + 5

What is the tangent of y= 9/(x^2+2),(1,3) ?
The tangent of y= 9/(x^2+2),(1,3) is y=-2x+5