What is the tangent of f(x)=2-6x^2,(-5,-148) ?

The tangent of f(x)=2-6x^2,(-5,-148) is y=60x+152

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tangent of f(x)=2-6x^2,(-5,-148)

tangent of

f (x) = 2 - 6 x^{2}, (- 5, - 148)

y = 60 x + 152

Solution steps

Find the slope of

f (x) = 2 - 6 x^{2} : \frac{df ( x )}{d x} = - 12 x

Compute the slope of

f (x) = 2 - 6 x^{2}

(- 5, - 148) : m = 60

Find the line with slope m=

60

and passing through

(- 5, - 148) : y = 60 x + 152

y = 60 x + 152

\frac{d}{d x} ((3 x^{4} - 6 x)^{3})

y = 4 e^{x} cos (x)

y = \frac{ln ( x )}{x}

\frac{\partial}{\partial x} (A \frac{x}{z})

\frac{d}{d x} (\frac{3 x - 4}{( 5 x + 2 ) ^{3}})

What is the tangent of f(x)=2-6x^2,(-5,-148) ?
The tangent of f(x)=2-6x^2,(-5,-148) is y=60x+152