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

The tangent of y=2x-3x^2 is y=(2-6a_{0})x+3a_{0}^2

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tangent of y=2x-3x^2

tangent of

y = 2 x - 3 x^{2}

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

Compute the slope of

y = 2 x - 3 x^{2}

x = a_{0} : m = 2 - 6 a_{0}

Find the line with slope m=

2 - 6 a_{0}

and passing through

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

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

\frac{d}{d x} (cos^{3} (5 x))

\int_{0}^{1} \frac{2 t}{( 2 - t ) ^{2}} d t

\int \frac{x ^{3}}{x ^{2} - 9} d x

\frac{9}{( 1 - x ) ^{2}}

n = 1 \sum \infty \frac{1}{n ^{5}}

What is the tangent of y=2x-3x^2 ?
The tangent of y=2x-3x^2 is y=(2-6a_{0})x+3a_{0}^2