What is the tangent of f(x)=4x^2+2x-1,\at x=2 ?

The tangent of f(x)=4x^2+2x-1,\at x=2 is y=18x-17

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tangent of f(x)=4x^2+2x-1,\at x=2

tangent of

f (x) = 4 x^{2} + 2 x - 1, at x = 2

y = 18 x - 17

Solution steps

Find the tangent point:

(2, 19)

Find the slope of

f (x) = 4 x^{2} + 2 x - 1 : \frac{df ( x )}{d x} = 8 x + 2

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 18

Find the line with slope m=

18

and passing through

(2, 19) : y = 18 x - 17

y = 18 x - 17

\frac{\partial}{\partial z} (\frac{1}{x y ^{2} z})

\int \frac{x ^{3}}{( 1 + x ^{2} ) ^{3}} d x

x \to \infty lim (\frac{3 x ^{2} - 2 x - 5}{x - 4})

\frac{\partial}{\partial y} (sin (y) - y sin (x))

x \to \infty lim (\frac{1}{3 x})

What is the tangent of f(x)=4x^2+2x-1,\at x=2 ?
The tangent of f(x)=4x^2+2x-1,\at x=2 is y=18x-17