What is the tangent of y=7x^2-5x+2 ?

The tangent of y=7x^2-5x+2 is y=(14a_{0}-5)x-7a_{0}^2+2

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tangent of y=7x^2-5x+2

tangent of

y = 7 x^{2} - 5 x + 2

y = (14 a_{0} - 5) x - 7 a_{0}^{2} + 2

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 7 a_{0}^{2} - 5 a_{0} + 2)

Find the slope of

y = 7 x^{2} - 5 x + 2 : \frac{d y}{d x} = 14 x - 5

Compute the slope of

y = 7 x^{2} - 5 x + 2

x = a_{0} : m = 14 a_{0} - 5

Find the line with slope m=

14 a_{0} - 5

and passing through

(a_{0}, 7 a_{0}^{2} - 5 a_{0} + 2) : y = (14 a_{0} - 5) x - 7 a_{0}^{2} + 2

y = (14 a_{0} - 5) x - 7 a_{0}^{2} + 2

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\frac{\partial}{\partial x} (21 - 3 x^{2} - 4 y^{2})

\frac{d}{d x} (7^{- 4 x + 1 + 2 x^{- 1}})

\frac{d}{d x} (x (- \frac{1}{10} x + 610))

What is the tangent of y=7x^2-5x+2 ?
The tangent of y=7x^2-5x+2 is y=(14a_{0}-5)x-7a_{0}^2+2