What is the tangent of f(x)=-3+(x+4)^2,\at x=-5 ?

The tangent of f(x)=-3+(x+4)^2,\at x=-5 is y=-2x-12

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tangent of f(x)=-3+(x+4)^2,\at x=-5

tangent of

f (x) = - 3 + (x + 4)^{2}, at x = - 5

y = - 2 x - 12

Solution steps

Find the tangent point:

(- 5, - 2)

Find the slope of

f (x) = - 3 + (x + 4)^{2} : \frac{df ( x )}{d x} = 2 x + 8

Compute the slope of

f (x) = - 3 + (x + 4)^{2}

x = - 5 : m = - 2

Find the line with slope m=

- 2

and passing through

(- 5, - 2) : y = - 2 x - 12

y = - 2 x - 12

\frac{2 \cdot ( s ^{2} + 1 )}{s ^{4}}

\frac{d}{d x} (e^{\frac{x ^{2}}{2 a}})

y^{^{'}} = y^{2} - 8

x \to 0 lim (\frac{\frac{5}{1 + x} - 5}{x})

\frac{d}{d x} (sin (2 x) d x)

What is the tangent of f(x)=-3+(x+4)^2,\at x=-5 ?
The tangent of f(x)=-3+(x+4)^2,\at x=-5 is y=-2x-12