What is the tangent of f(x)=5x^2-6x+1,\at x=2 ?

The tangent of f(x)=5x^2-6x+1,\at x=2 is y=14x-19

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tangent of f(x)=5x^2-6x+1,\at x=2

tangent of

f (x) = 5 x^{2} - 6 x + 1, at x = 2

y = 14 x - 19

Solution steps

Find the tangent point:

(2, 9)

Find the slope of

f (x) = 5 x^{2} - 6 x + 1 : \frac{df ( x )}{d x} = 10 x - 6

Compute the slope of

f (x) = 5 x^{2} - 6 x + 1

x = 2 : m = 14

Find the line with slope m=

14

and passing through

(2, 9) : y = 14 x - 19

y = 14 x - 19

(sin (θ)) \frac{d y}{d θ} = cos (θ)

\int (x^{3} - 1) e^{- \frac{x}{3}} d x

\frac{d}{d x} (- \frac{120}{x ^{6}})

\frac{d}{d z} (x y)

x^{3} y^{^{'}} + 2 x^{2} y = - 4 cos (2 x) y^{- \frac{1}{2}}

What is the tangent of f(x)=5x^2-6x+1,\at x=2 ?
The tangent of f(x)=5x^2-6x+1,\at x=2 is y=14x-19