What is the tangent of f(x)=8x-3x^2 ?

The tangent of f(x)=8x-3x^2 is y=(8-6a_{0})x+3a_{0}^2

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tangent of f(x)=8x-3x^2

tangent of

f (x) = 8 x - 3 x^{2}

y = (8 - 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}, 8 a_{0} - 3 a_{0}^{2})

Find the slope of

f (x) = 8 x - 3 x^{2} : \frac{df ( x )}{d x} = 8 - 6 x

Compute the slope of

f (x) = 8 x - 3 x^{2}

x = a_{0} : m = 8 - 6 a_{0}

Find the line with slope m=

8 - 6 a_{0}

and passing through

(a_{0}, 8 a_{0} - 3 a_{0}^{2}) : y = (8 - 6 a_{0}) x + 3 a_{0}^{2}

y = (8 - 6 a_{0}) x + 3 a_{0}^{2}

\int_{- 1}^{2} \frac{1}{x ^{5}} d x

- \frac{4}{x ^{3}}

\int 2 (x + y) d x

\frac{d ^{2}}{d x ^{2}} (x sin (x))

\frac{\partial}{\partial x} (cos (y^{2} - 2 x))

What is the tangent of f(x)=8x-3x^2 ?
The tangent of f(x)=8x-3x^2 is y=(8-6a_{0})x+3a_{0}^2