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

The tangent of f(x)=4x+3x^2,\at x=2 is y=16x-12

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

tangent of

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

y = 16 x - 12

Solution steps

Find the tangent point:

(2, 20)

Find the slope of

f (x) = 4 x + 3 x^{2} : \frac{df ( x )}{d x} = 4 + 6 x

Compute the slope of

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

x = 2 : m = 16

Find the line with slope m=

16

and passing through

(2, 20) : y = 16 x - 12

y = 16 x - 12

\frac{\partial}{\partial x} (x^{2} z^{2})

e^{(c o s (t) + l n (t))}

\frac{\partial}{\partial x} (e^{t - x} + e^{w - 1})

\int cos (6 x) cos (5 x) d x

g (t) = \frac{t - t}{t ^{\frac{1}{5}}}

What is the tangent of f(x)=4x+3x^2,\at x=2 ?
The tangent of f(x)=4x+3x^2,\at x=2 is y=16x-12