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

The tangent of f(x)=4x^2+2x,(-2,12) is y=-14x-16

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

tangent of

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

y = - 14 x - 16

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

(- 2, 12) : m = - 14

Find the line with slope m=

- 14

and passing through

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

y = - 14 x - 16

\int \frac{x ^{7}}{2} - \frac{1}{x ^{5}} d x

n \to \infty lim (sin (2 n))

e^{- 3 x}, 0

n = 2 \sum \infty x^{n}

\frac{d y}{d x} = - \frac{sin ( x )}{y}

What is the tangent of f(x)=4x^2+2x,(-2,12) ?
The tangent of f(x)=4x^2+2x,(-2,12) is y=-14x-16