What is the tangent of f(x)=3-4x^2,(2,-13) ?

The tangent of f(x)=3-4x^2,(2,-13) is y=-16x+19

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tangent of f(x)=3-4x^2,(2,-13)

tangent of

f (x) = 3 - 4 x^{2}, (2, - 13)

y = - 16 x + 19

Solution steps

Find the slope of

f (x) = 3 - 4 x^{2} : \frac{df ( x )}{d x} = - 8 x

Compute the slope of

f (x) = 3 - 4 x^{2}

(2, - 13) : m = - 16

Find the line with slope m=

- 16

and passing through

(2, - 13) : y = - 16 x + 19

y = - 16 x + 19

ln (\frac{1}{x})

\int \frac{x + 1}{2 + 4 x ^{2}} d x

\frac{d}{d x} (x^{3} cot (x))

\int \frac{( sin ( w ) )}{w} d w

\frac{( s - 2 )}{( s - 2 ) ^{2} - 3}

What is the tangent of f(x)=3-4x^2,(2,-13) ?
The tangent of f(x)=3-4x^2,(2,-13) is y=-16x+19