What is the tangent of f(x)=x^2+1,(-5,26) ?

The tangent of f(x)=x^2+1,(-5,26) is y=-10x-24

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tangent of f(x)=x^2+1,(-5,26)

tangent of

f (x) = x^{2} + 1, (- 5, 26)

y = - 10 x - 24

Solution steps

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

(- 5, 26) : m = - 10

Find the line with slope m=

- 10

and passing through

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

y = - 10 x - 24

\frac{\partial}{\partial x} (3 x^{2} - 6 x - 9)

\int \frac{1}{4 x + 1} d x

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

sin (7 x)

x^{3} - 2 x^{2} + 4, 3 x^{2} + 5 x - 21

What is the tangent of f(x)=x^2+1,(-5,26) ?
The tangent of f(x)=x^2+1,(-5,26) is y=-10x-24