What is the tangent of f(x)=x^2+4,(-4,20) ?

The tangent of f(x)=x^2+4,(-4,20) is y=-8x-12

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tangent of f(x)=x^2+4,(-4,20)

tangent of

f (x) = x^{2} + 4, (- 4, 20)

y = - 8 x - 12

Solution steps

Find the slope of

f (x) = x^{2} + 4 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = x^{2} + 4

(- 4, 20) : m = - 8

Find the line with slope m=

- 8

and passing through

(- 4, 20) : y = - 8 x - 12

y = - 8 x - 12

\frac{d}{d u} (2 uv)

\frac{d}{d x} (ln ∣ coth (5 x) ∣)

y^{^{'}} + e^{y} t = e^{y} sin (t)

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

\frac{\partial}{\partial x} (- 2 cos (4 x))

What is the tangent of f(x)=x^2+4,(-4,20) ?
The tangent of f(x)=x^2+4,(-4,20) is y=-8x-12