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

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

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tangent of y=3x^2-5x

tangent of

y = 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

y = 3 x^{2} - 5 x : \frac{d y}{d x} = 6 x - 5

Compute the slope of

y = 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}

\frac{\partial}{\partial x} (2 x + 2 y + z)

\frac{d}{d x} (x^{3} y^{4})

f (x) = \frac{5 x - 9}{x + 1}, at x = 0

f (x) = \frac{x ^{2}}{x - 1}

\int 4 x 1 - x^{2} d x

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