What is the tangent of f(x)=3x^2-4x+1,\at x=2 ?

The tangent of f(x)=3x^2-4x+1,\at x=2 is y=8x-11

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tangent of f(x)=3x^2-4x+1,\at x=2

tangent of

f (x) = 3 x^{2} - 4 x + 1, at x = 2

y = 8 x - 11

Solution steps

Find the tangent point:

(2, 5)

Find the slope of

f (x) = 3 x^{2} - 4 x + 1 : \frac{df ( x )}{d x} = 6 x - 4

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 = 8

Find the line with slope m=

8

and passing through

(2, 5) : y = 8 x - 11

y = 8 x - 11

\int (2 u^{2} - 2 u^{- 1}) d u

3 y^{^{''}} + 2 y^{^{'}} + y = 0

\frac{\partial}{\partial y} (\frac{y}{( x ^{2} + y ^{2} )})

f (x) = 4 x^{6} - 7 x^{5}

5 x sin (π x)

What is the tangent of f(x)=3x^2-4x+1,\at x=2 ?
The tangent of f(x)=3x^2-4x+1,\at x=2 is y=8x-11