What is the tangent of f(x)=6x^2+8x-15,\at x=2 ?

The tangent of f(x)=6x^2+8x-15,\at x=2 is y=32x-39

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tangent of f(x)=6x^2+8x-15,\at x=2

tangent of

f (x) = 6 x^{2} + 8 x - 15, at x = 2

y = 32 x - 39

Solution steps

Find the tangent point:

(2, 25)

Find the slope of

f (x) = 6 x^{2} + 8 x - 15 : \frac{df ( x )}{d x} = 12 x + 8

Compute the slope of

f (x) = 6 x^{2} + 8 x - 15

x = 2 : m = 32

Find the line with slope m=

32

and passing through

(2, 25) : y = 32 x - 39

y = 32 x - 39

\frac{\partial}{\partial y} (e^{- x^{2} - y^{2}})

x \to 3 lim (\frac{tan ( x - 3 )}{3 e ^{x - 3} - x})

\frac{1}{x ^{2} + 4}

\frac{d}{d x} (x^{7} ln (8 x))

\frac{d}{d x} (\frac{tan ( x )}{x ^{2}})

What is the tangent of f(x)=6x^2+8x-15,\at x=2 ?
The tangent of f(x)=6x^2+8x-15,\at x=2 is y=32x-39