What is the tangent of f(x)=3x-2x^2 ?

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

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tangent of f(x)=3x-2x^2

tangent of

f (x) = 3 x - 2 x^{2}

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

f (x) = 3 x - 2 x^{2} : \frac{df ( x )}{d x} = 3 - 4 x

Compute the slope of

f (x) = 3 x - 2 x^{2}

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

Find the line with slope m=

3 - 4 a_{0}

and passing through

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

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

\frac{d y}{d x} - y = e^{x}

n = 0 \sum \infty 9 (\frac{2}{3})^{n}

\frac{d y}{d t} = 9 y - y^{2} - \frac{81}{4}

\frac{d}{d a} (sin^{4} (a) - cos^{4} (a) + 2 cos^{2} (a))

\int \frac{sin ( 2 x )}{1 + cos ^{4} ( x )} d x

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