What is the tangent of y=(3x)/(x^2-6),(3,3) ?

The tangent of y=(3x)/(x^2-6),(3,3) is y=-5x+18

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tangent of y=(3x)/(x^2-6),(3,3)

tangent of

y = \frac{3 x}{x ^{2} - 6}, (3, 3)

y = - 5 x + 18

Solution steps

Find the slope of

y = \frac{3 x}{x ^{2} - 6} : \frac{d y}{d x} = \frac{3 ( - x ^{2} - 6 )}{( x ^{2} - 6 ) ^{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 = - 5

Find the line with slope m=

- 5

and passing through

(3, 3) : y = - 5 x + 18

y = - 5 x + 18

x \to 1 - lim (\frac{x - 5}{ln ( x )})

\int x^{\frac{1}{2}} (- 6 + 6 x) d x

\int_{0}^{3} \frac{1}{9 + t ^{2}} d t

n = 0 \sum \infty (0.8)^{n}

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

What is the tangent of y=(3x)/(x^2-6),(3,3) ?
The tangent of y=(3x)/(x^2-6),(3,3) is y=-5x+18