What is the tangent of f(x)=x^2+6x-16,\at x=2 ?

The tangent of f(x)=x^2+6x-16,\at x=2 is y=10x-20

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tangent of f(x)=x^2+6x-16,\at x=2

tangent of

f (x) = x^{2} + 6 x - 16, at x = 2

y = 10 x - 20

Solution steps

Find the tangent point:

(2, 0)

Find the slope of

f (x) = x^{2} + 6 x - 16 : \frac{df ( x )}{d x} = 2 x + 6

Compute the slope of

f (x) = x^{2} + 6 x - 16

x = 2 : m = 10

Find the line with slope m=

10

and passing through

(2, 0) : y = 10 x - 20

y = 10 x - 20

x \to 0 lim (\frac{h ( x )}{g ( x )})

(8 x - x^{2})^{3}

e^{t} \cdot cos (t) + j \cdot e^{t} \cdot sin (t)

e^{x} d x = 2 y d y

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

What is the tangent of f(x)=x^2+6x-16,\at x=2 ?
The tangent of f(x)=x^2+6x-16,\at x=2 is y=10x-20