What is the tangent of f(x)=3x^2-5x ?

The tangent of f(x)=3x^2-5x is y=(6a_{0}-5)x-3a_{0}^2

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tangent of f(x)=3x^2-5x

tangent of

f (x) = 3 x^{2} - 5 x

y = (6 a_{0} - 5) x - 3 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} - 5 a_{0})

Find the slope of

f (x) = 3 x^{2} - 5 x : \frac{df ( x )}{d x} = 6 x - 5

Compute the slope of

f (x) = 3 x^{2} - 5 x

x = a_{0} : m = 6 a_{0} - 5

Find the line with slope m=

6 a_{0} - 5

and passing through

(a_{0}, 3 a_{0}^{2} - 5 a_{0}) : y = (6 a_{0} - 5) x - 3 a_{0}^{2}

y = (6 a_{0} - 5) x - 3 a_{0}^{2}

f (x) = - \frac{4}{3} x^{- \frac{5}{2}}, at x = 4

\frac{d}{d x} (x (x - 1) (x - 2))

- 4 x^{2} + 9, x + 7

y^{^{'}} + 5 y = t e^{(4 t)}

What is the tangent of f(x)=3x^2-5x ?
The tangent of f(x)=3x^2-5x is y=(6a_{0}-5)x-3a_{0}^2