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

The tangent of f(x)=4x^2+2,\at x=-2 is y=-16x-14

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

tangent of

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

y = - 16 x - 14

Solution steps

Find the tangent point:

(- 2, 18)

Find the slope of

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

Compute the slope of

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

x = - 2 : m = - 16

Find the line with slope m=

- 16

and passing through

(- 2, 18) : y = - 16 x - 14

y = - 16 x - 14

\frac{d ^{2} y}{d t ^{2}} = 1 - e^{2 t}

\frac{d}{d s} (ln (3))

\int \frac{1}{9 + 2 x} d x

2 y = 3 x, 2 y + 3 x = 6, y = - 3

f (x) = \frac{2 x ^{2} - 5 x}{3 x + 5}

What is the tangent of f(x)=4x^2+2,\at x=-2 ?
The tangent of f(x)=4x^2+2,\at x=-2 is y=-16x-14