What is the tangent of f(x)=x^2+1,\at x=-5 ?

The tangent of f(x)=x^2+1,\at x=-5 is y=-10x-24

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tangent of f(x)=x^2+1,\at x=-5

tangent of

f (x) = x^{2} + 1, at x = - 5

y = - 10 x - 24

Solution steps

Find the tangent point:

(- 5, 26)

Find the slope of

f (x) = x^{2} + 1 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = x^{2} + 1

x = - 5 : m = - 10

Find the line with slope m=

- 10

and passing through

(- 5, 26) : y = - 10 x - 24

y = - 10 x - 24

\int_{0}^{1} cos (π x) d x

\frac{d ^{4}}{d x ^{4}} (sin (x))

x \to 3 lim (2 x^{3} - 4 x - 2)

\int x^{35} sin (x^{36}) d x

- 4 cos (5 t)

What is the tangent of f(x)=x^2+1,\at x=-5 ?
The tangent of f(x)=x^2+1,\at x=-5 is y=-10x-24