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

The tangent of f(x)=x^2,\at x=-5 is y=-10x-25

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

tangent of

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

y = - 10 x - 25

Solution steps

Find the tangent point:

(- 5, 25)

Find the slope of

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

Compute the slope of

f (x) = x^{2}

x = - 5 : m = - 10

Find the line with slope m=

- 10

and passing through

(- 5, 25) : y = - 10 x - 25

y = - 10 x - 25

(\frac{1}{s}) (\frac{10}{s + 5})

\frac{3 x + 4}{2 x - 1}

\frac{d}{d x} (4 x^{- \frac{1}{3}})

y (6 y^{2} - x - 1) d x + 2 x d y = 0

\frac{d}{d x} (\frac{sin ( x )}{cos ^{3} ( x )})

What is the tangent of f(x)=x^2,\at x=-5 ?
The tangent of f(x)=x^2,\at x=-5 is y=-10x-25