What is the tangent of y=x^3-2x^2+5 ?

The tangent of y=x^3-2x^2+5 is y=(3a_{0}^2-4a_{0})x-2a_{0}^3+2a_{0}^2+5

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tangent of y=x^3-2x^2+5

tangent of

y = x^{3} - 2 x^{2} + 5

y = (3 a_{0}^{2} - 4 a_{0}) x - 2 a_{0}^{3} + 2 a_{0}^{2} + 5

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{3} - 2 a_{0}^{2} + 5)

Find the slope of

y = x^{3} - 2 x^{2} + 5 : \frac{d y}{d x} = 3 x^{2} - 4 x

Compute the slope of

y = x^{3} - 2 x^{2} + 5

x = a_{0} : m = 3 a_{0}^{2} - 4 a_{0}

Find the line with slope m=

3 a_{0}^{2} - 4 a_{0}

and passing through

(a_{0}, a_{0}^{3} - 2 a_{0}^{2} + 5) : y = (3 a_{0}^{2} - 4 a_{0}) x - 2 a_{0}^{3} + 2 a_{0}^{2} + 5

y = (3 a_{0}^{2} - 4 a_{0}) x - 2 a_{0}^{3} + 2 a_{0}^{2} + 5

\int r ln (r) d r

x \to 1 lim (\frac{x ^{8} - 1}{x - 1})

n = 1 \sum \infty \frac{( 3 ^{n - 1} )}{9 ^{n}}

\frac{d}{d x} (4 x 16 - x^{2})

\int_{1}^{2} 2 x x^{2} + 1 d x

What is the tangent of y=x^3-2x^2+5 ?
The tangent of y=x^3-2x^2+5 is y=(3a_{0}^2-4a_{0})x-2a_{0}^3+2a_{0}^2+5