What is the tangent of f(x)=3x^2-5,\at x=2 ?

The tangent of f(x)=3x^2-5,\at x=2 is y=12x-17

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tangent of f(x)=3x^2-5,\at x=2

tangent of

f (x) = 3 x^{2} - 5, at x = 2

y = 12 x - 17

Solution steps

Find the tangent point:

(2, 7)

Find the slope of

f (x) = 3 x^{2} - 5 : \frac{df ( x )}{d x} = 6 x

Compute the slope of

f (x) = 3 x^{2} - 5

x = 2 : m = 12

Find the line with slope m=

12

and passing through

(2, 7) : y = 12 x - 17

y = 12 x - 17

x \to 3 π + lim (cot (x))

\int - 2 x cos (6 x) d x

\frac{\partial}{\partial x} (ln (\frac{x y}{y + z}))

\frac{6}{s ( s ^{2} + 2 s + 6 )}

f (x) = \frac{1 + x ^{2}}{x ^{2} - 1}

What is the tangent of f(x)=3x^2-5,\at x=2 ?
The tangent of f(x)=3x^2-5,\at x=2 is y=12x-17