What is the tangent of f(x)=-3-4x^2,(5,-103) ?

The tangent of f(x)=-3-4x^2,(5,-103) is y=-40x+97

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tangent of f(x)=-3-4x^2,(5,-103)

tangent of

f (x) = - 3 - 4 x^{2}, (5, - 103)

y = - 40 x + 97

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}

(5, - 103) : m = - 40

Find the line with slope m=

- 40

and passing through

(5, - 103) : y = - 40 x + 97

y = - 40 x + 97

f (x) = 2 x^{3} - x^{2} + 1, at x = 2

sin (x) - cos (x)

x \frac{d y}{d x} + y = x^{4} ln (x)

x \to 2 + lim (\frac{x + 2}{x ^{2} - 4})

5 + 2 x

What is the tangent of f(x)=-3-4x^2,(5,-103) ?
The tangent of f(x)=-3-4x^2,(5,-103) is y=-40x+97