What is the tangent of y=6x^2-3x,\at 1,5 ?

The tangent of y=6x^2-3x,\at 1,5 is y=(12a_{0}-3)x-6a_{0}^2

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tangent of y=6x^2-3x,\at 1,5

tangent of

y = 6 x^{2} - 3 x, at 1, 5

y = (12 a_{0} - 3) x - 6 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 6 a_{0}^{2} - 3 a_{0})

Find the slope of

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

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 = 12 a_{0} - 3

Find the line with slope m=

12 a_{0} - 3

and passing through

(a_{0}, 6 a_{0}^{2} - 3 a_{0}) : y = (12 a_{0} - 3) x - 6 a_{0}^{2}

y = (12 a_{0} - 3) x - 6 a_{0}^{2}

(8.1) (4.1)

\frac{\partial}{\partial y} (x^{4} + 2 x^{2} y)

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

n = 0 \sum \infty n \cdot sin (n \cdot π)

\int \frac{x ^{2} - 5 x + 6}{x - 2} d x

What is the tangent of y=6x^2-3x,\at 1,5 ?
The tangent of y=6x^2-3x,\at 1,5 is y=(12a_{0}-3)x-6a_{0}^2