What is the tangent of y=6x^2-6x+3,(1,3) ?

The tangent of y=6x^2-6x+3,(1,3) is y=6x-3

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tangent of y=6x^2-6x+3,(1,3)

tangent of

y = 6 x^{2} - 6 x + 3, (1, 3)

y = 6 x - 3

Solution steps

Find the slope of

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

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 = 6

Find the line with slope m=

6

and passing through

(1, 3) : y = 6 x - 3

y = 6 x - 3

\int 9 y d y

\int \frac{2}{3} x^{- \frac{1}{3}} d x

n \to 0 lim ((1 + n)^{\frac{1}{n}})

f (x) = \frac{7}{x ^{3}}

y = 0, y = ln (x), y = 0, [1, 2]

What is the tangent of y=6x^2-6x+3,(1,3) ?
The tangent of y=6x^2-6x+3,(1,3) is y=6x-3