What is the tangent of f(x)=x^2-5,(-4,11) ?

The tangent of f(x)=x^2-5,(-4,11) is y=-8x-21

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tangent of f(x)=x^2-5,(-4,11)

tangent of

f (x) = x^{2} - 5, (- 4, 11)

y = - 8 x - 21

Solution steps

Find the slope of

f (x) = x^{2} - 5 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = x^{2} - 5

(- 4, 11) : m = - 8

Find the line with slope m=

- 8

and passing through

(- 4, 11) : y = - 8 x - 21

y = - 8 x - 21

\int_{0}^{10} 1000 t e^{- 0.5 t} d t

\frac{d}{d θ} (e^{i θ})

\int 2 cos^{2} (θ) - \frac{1}{2} d θ

\frac{\partial}{\partial y} ((x^{2} - 2 x) cos (y))

\frac{\partial ^{2}}{\partial x ^{2}} (ln (2 + x^{2} y^{2}))

What is the tangent of f(x)=x^2-5,(-4,11) ?
The tangent of f(x)=x^2-5,(-4,11) is y=-8x-21