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

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

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

tangent of

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

y = 4 x - 2

Solution steps

Find the slope of

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

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 = 4

Find the line with slope m=

4

and passing through

(1, 2) : y = 4 x - 2

y = 4 x - 2

\frac{( 4 x - 2 ) ^{4}}{( - 5 x + 1 ) ^{5}}

f (x) = (\frac{1}{x - 3})^{2}

x^{3} + 2 x^{2} + 1

\int \frac{x ^{3}}{3} - \frac{3}{x ^{3}} d x

\int \frac{x ^{2}}{x ^{2} + 9} d x

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