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

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

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

tangent of

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

y = - 12 x + 1

Solution steps

Find the slope of

y = 3 x^{2} - 12 x + 1 : \frac{d y}{d x} = 6 x - 12

Compute the slope of

y = 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

f (x) = x^{2} + 2

\frac{d}{d x} (\frac{x ^{2} - 3}{x + 1})

x = 1 \sum \infty x^{3} e^{- x^{2}}

y^{^{''}} + 64 y = 0, y (0) = 2, y^{^{'}} (0) = 1

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