What is the tangent of y=x^2-8x+9,(1,2) ?

The tangent of y=x^2-8x+9,(1,2) is y=-6x+8

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tangent of y=x^2-8x+9,(1,2)

tangent of

y = x^{2} - 8 x + 9, (1, 2)

y = - 6 x + 8

Solution steps

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

(1, 2) : m = - 6

Find the line with slope m=

- 6

and passing through

(1, 2) : y = - 6 x + 8

y = - 6 x + 8

\frac{d}{d x} (u e^{- x})

\int 361 - x^{2} d x

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

\int (\frac{e ^{x}}{3} + 2 x) d x

\int tan^{5} (x) \cdot sec^{3} (x) d x

What is the tangent of y=x^2-8x+9,(1,2) ?
The tangent of y=x^2-8x+9,(1,2) is y=-6x+8