What is the tangent of x^2-8x+9 ?

The tangent of x^2-8x+9 is y=(2a_{0}-8)x-a_{0}^2+9

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tangent of x^2-8x+9

tangent of

x^{2} - 8 x + 9

y = (2 a_{0} - 8) x - a_{0}^{2} + 9

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{2} - 8 a_{0} + 9)

Find the slope of

y = x^{2} - 8 x + 9 : \frac{d y}{d x} = 2 x - 8

Compute the slope of

y = x^{2} - 8 x + 9

x = a_{0} : m = 2 a_{0} - 8

Find the line with slope m=

2 a_{0} - 8

and passing through

(a_{0}, a_{0}^{2} - 8 a_{0} + 9) : y = (2 a_{0} - 8) x - a_{0}^{2} + 9

y = (2 a_{0} - 8) x - a_{0}^{2} + 9

x^{2} - 2 x + 2, (0, 1)

\int (2 y + 3)^{2} d y

\frac{d}{d t} (e^{- 3 t})

x \to \infty lim (3 x + \frac{1}{2})

\int \frac{2}{x ^{2} - 2 x} d x

What is the tangent of x^2-8x+9 ?
The tangent of x^2-8x+9 is y=(2a_{0}-8)x-a_{0}^2+9