What is the tangent of f(x)=3x^2-12x+1,(0,1) ?

The tangent of f(x)=3x^2-12x+1,(0,1) is y=-12x+1

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tangent of f(x)=3x^2-12x+1,(0,1)

tangent of

f (x) = 3 x^{2} - 12 x + 1, (0, 1)

y = - 12 x + 1

Solution steps

Find the slope of

f (x) = 3 x^{2} - 12 x + 1 : \frac{df ( x )}{d x} = 6 x - 12

Compute the slope of

f (x) = 3 x^{2} - 12 x + 1

(0, 1) : m = - 12

Find the line with slope m=

- 12

and passing through

(0, 1) : y = - 12 x + 1

y = - 12 x + 1

\frac{d}{d x} (\frac{x ^{3}}{3} - 2 x^{2} - 5 x)

f (x) = x^{2} - 4, at x = 5

n = 0 \sum \infty 3^{- 2 n}

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

15 - x^{2}, - 6 x - 1, [- 4, - 1]

What is the tangent of f(x)=3x^2-12x+1,(0,1) ?
The tangent of f(x)=3x^2-12x+1,(0,1) is y=-12x+1