What is the tangent of f(x)=4x^2+12x+1,\at x=-2 ?

The tangent of f(x)=4x^2+12x+1,\at x=-2 is y=-4x-15

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tangent of f(x)=4x^2+12x+1,\at x=-2

tangent of

f (x) = 4 x^{2} + 12 x + 1, at x = - 2

y = - 4 x - 15

Solution steps

Find the tangent point:

(- 2, - 7)

Find the slope of

f (x) = 4 x^{2} + 12 x + 1 : \frac{df ( x )}{d x} = 8 x + 12

Compute the slope of

f (x) = 4 x^{2} + 12 x + 1

x = - 2 : m = - 4

Find the line with slope m=

- 4

and passing through

(- 2, - 7) : y = - 4 x - 15

y = - 4 x - 15

\frac{d ^{2}}{d x ^{2}} (sin (3 x^{2}))

\frac{d}{d x} (x^{r (x)})

\frac{d}{d x} (30)

- x^{2} + 3 x, x - 3

\frac{d y}{d x} = \frac{( ( 1 + y ) ^{2} )}{( 1 + x ) ^{3}}

What is the tangent of f(x)=4x^2+12x+1,\at x=-2 ?
The tangent of f(x)=4x^2+12x+1,\at x=-2 is y=-4x-15