What is the tangent of 12x^2-8x-4 ?

The tangent of 12x^2-8x-4 is y=(24a_{0}-8)x-12a_{0}^2-4

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tangent of 12x^2-8x-4

tangent of

12 x^{2} - 8 x - 4

y = (24 a_{0} - 8) x - 12 a_{0}^{2} - 4

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 12 a_{0}^{2} - 8 a_{0} - 4)

Find the slope of

y = 12 x^{2} - 8 x - 4 : \frac{d y}{d x} = 24 x - 8

Compute the slope of

y = 12 x^{2} - 8 x - 4

x = a_{0} : m = 24 a_{0} - 8

Find the line with slope m=

24 a_{0} - 8

and passing through

(a_{0}, 12 a_{0}^{2} - 8 a_{0} - 4) : y = (24 a_{0} - 8) x - 12 a_{0}^{2} - 4

y = (24 a_{0} - 8) x - 12 a_{0}^{2} - 4

\int \frac{1}{e ^{x} - 1} d x

y^{^{''}} - 4 y^{^{'}} - 12 y = sin (2 x)

f (x) = \frac{2}{x - 2}, at x = - 4

x \to 0 lim (\frac{cos ( x ) - x}{x})

y = x^{3}, y = x^{\frac{1}{2}}

What is the tangent of 12x^2-8x-4 ?
The tangent of 12x^2-8x-4 is y=(24a_{0}-8)x-12a_{0}^2-4