What is the tangent of f(x)=x^2+4,(-5,29) ?

The tangent of f(x)=x^2+4,(-5,29) is y=-10x-21

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tangent of f(x)=x^2+4,(-5,29)

tangent of

f (x) = x^{2} + 4, (- 5, 29)

y = - 10 x - 21

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

(- 5, 29) : m = - 10

Find the line with slope m=

- 10

and passing through

(- 5, 29) : y = - 10 x - 21

y = - 10 x - 21

\frac{d}{d x} (x 2 x - 3)

y = x^{2} - 8 x + 5

x \to 5 lim (1)

\frac{\partial}{\partial x} (ln (4 x + 3 y))

y^{^{'}} + 2 y = 4 cos (2 x)

What is the tangent of f(x)=x^2+4,(-5,29) ?
The tangent of f(x)=x^2+4,(-5,29) is y=-10x-21