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

The tangent of f(x)=4x^2+2x,(-1,2) is y=-6x-4

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

tangent of

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

y = - 6 x - 4

Solution steps

Find the slope of

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

Compute the slope of

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

(- 1, 2) : m = - 6

Find the line with slope m=

- 6

and passing through

(- 1, 2) : y = - 6 x - 4

y = - 6 x - 4

- \frac{1}{2 x ^{2}} + \frac{32}{5 x ^{5}}

\int \frac{1}{x ^{- \frac{1}{2}}} d x

\frac{d}{d x} (x (4 x - 12)^{3})

\int - \frac{4000}{x ^{2}} d x

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

What is the tangent of f(x)=4x^2+2x,(-1,2) ?
The tangent of f(x)=4x^2+2x,(-1,2) is y=-6x-4