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

The tangent of y=6x^2-x^3,(1,5) is y=9x-4

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tangent of y=6x^2-x^3,(1,5)

tangent of

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

y = 9 x - 4

Solution steps

Find the slope of

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

Compute the slope of

y = 6 x^{2} - x^{3}

(1, 5) : m = 9

Find the line with slope m=

9

and passing through

(1, 5) : y = 9 x - 4

y = 9 x - 4

n = 0 \sum \infty \frac{1}{1 !}

\int 5^{x + 2} d x

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

(1, 3), (3, - 2)

\int - cos (t) - sin (t) d t

What is the tangent of y=6x^2-x^3,(1,5) ?
The tangent of y=6x^2-x^3,(1,5) is y=9x-4