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

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

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

tangent of

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

y = 2 x - 3

Solution steps

Find the slope of

f (x) = 3 x^{2} - 4 x : \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 = 2

Find the line with slope m=

2

and passing through

(1, - 1) : y = 2 x - 3

y = 2 x - 3

- 3 cos (x)

y = tan (x), y = 2 sin (x)

\int cot^{4} (2 t) d t

\frac{d}{d x} (\frac{( x ^{2} - 4 ) ^{3}}{x ^{2} + 2})

f (X) = ln (ln (ln (tan (X))))

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