What is the tangent of y=-2-7x^2,(3,-65) ?

The tangent of y=-2-7x^2,(3,-65) is y=-42x+61

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tangent of y=-2-7x^2,(3,-65)

tangent of

y = - 2 - 7 x^{2}, (3, - 65)

y = - 42 x + 61

Solution steps

Find the slope of

y = - 2 - 7 x^{2} : \frac{d y}{d x} = - 14 x

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 = - 42

Find the line with slope m=

- 42

and passing through

(3, - 65) : y = - 42 x + 61

y = - 42 x + 61

\frac{d}{d t} (t^{- 1})

(10 + 9 x)^{4} \frac{d y}{d x} = (6 + y)^{2}

x \to - \infty lim (x^{8} e^{x})

\frac{d}{d x} (\frac{( x ^{2} + 1 ) ( x + 3 ) ^{\frac{1}{2}}}{x - 1})

y^{^{'}} (x^{2} + 1) + 5 x (y - 1) = 0, y (0) = 7

What is the tangent of y=-2-7x^2,(3,-65) ?
The tangent of y=-2-7x^2,(3,-65) is y=-42x+61