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

The tangent of f(x)=4x^2+12x+3,\at x=-2 is y=-4x-13

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

tangent of

f (x) = 4 x^{2} + 12 x + 3, at x = - 2

y = - 4 x - 13

Solution steps

Find the tangent point:

(- 2, - 5)

Find the slope of

f (x) = 4 x^{2} + 12 x + 3 : \frac{df ( x )}{d x} = 8 x + 12

Compute the slope of

f (x) = 4 x^{2} + 12 x + 3

x = - 2 : m = - 4

Find the line with slope m=

- 4

and passing through

(- 2, - 5) : y = - 4 x - 13

y = - 4 x - 13

\int 5 x^{4} - 8 x^{5} d x

y^{^{''}} = a \cdot y

(x^{2} - 6 x + 10) 25 - x^{2}

4 sec (x) tan (x)

y^{^{'}} = 3 x + y, y (0) = 3

What is the tangent of f(x)=4x^2+12x+3,\at x=-2 ?
The tangent of f(x)=4x^2+12x+3,\at x=-2 is y=-4x-13