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

The tangent of f(x)=2x^2+x+1,\at x=2 is y=9x-7

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tangent of f(x)=2x^2+x+1,\at x=2

tangent of

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

y = 9 x - 7

Solution steps

Find the tangent point:

(2, 11)

Find the slope of

f (x) = 2 x^{2} + x + 1 : \frac{df ( x )}{d x} = 4 x + 1

Compute the slope of

f (x) = 2 x^{2} + x + 1

x = 2 : m = 9

Find the line with slope m=

9

and passing through

(2, 11) : y = 9 x - 7

y = 9 x - 7

\frac{\partial}{\partial x} (2 arctan (\frac{y}{x}))

y = 7 sec (x), at x = \frac{π}{3}

\frac{d}{d t} (arctan (\frac{t}{2}))

x^{2} + 5

y = \frac{sec ( x )}{1 + sec ( x )}

What is the tangent of f(x)=2x^2+x+1,\at x=2 ?
The tangent of f(x)=2x^2+x+1,\at x=2 is y=9x-7