What is the tangent of 5x^2-4x+5 ?

The tangent of 5x^2-4x+5 is y=(10a_{0}-4)x-5a_{0}^2+5

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tangent of 5x^2-4x+5

tangent of

5 x^{2} - 4 x + 5

y = (10 a_{0} - 4) x - 5 a_{0}^{2} + 5

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 5 a_{0}^{2} - 4 a_{0} + 5)

Find the slope of

y = 5 x^{2} - 4 x + 5 : \frac{d y}{d x} = 10 x - 4

Compute the slope of

y = 5 x^{2} - 4 x + 5

x = a_{0} : m = 10 a_{0} - 4

Find the line with slope m=

10 a_{0} - 4

and passing through

(a_{0}, 5 a_{0}^{2} - 4 a_{0} + 5) : y = (10 a_{0} - 4) x - 5 a_{0}^{2} + 5

y = (10 a_{0} - 4) x - 5 a_{0}^{2} + 5

x \to \infty lim (x e^{x})

\frac{d y}{d x} = \frac{9 x}{7 y}

\frac{d}{d y} (ln (x y))

\frac{\partial}{\partial x} (5 cos (x))

\int 2 x (x^{2} + 1)^{- 5} d x

What is the tangent of 5x^2-4x+5 ?
The tangent of 5x^2-4x+5 is y=(10a_{0}-4)x-5a_{0}^2+5