What is the tangent of 8x^2-7x,\at 9 ?

The tangent of 8x^2-7x,\at 9 is y=(16a_{0}-7)x-8a_{0}^2

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tangent of 8x^2-7x,\at 9

tangent of

8 x^{2} - 7 x, at 9

y = (16 a_{0} - 7) x - 8 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 8 a_{0}^{2} - 7 a_{0})

Find the slope of

y = 8 x^{2} - 7 x : \frac{d y}{d x} = 16 x - 7

Compute the slope of

y = 8 x^{2} - 7 x

x = a_{0} : m = 16 a_{0} - 7

Find the line with slope m=

16 a_{0} - 7

and passing through

(a_{0}, 8 a_{0}^{2} - 7 a_{0}) : y = (16 a_{0} - 7) x - 8 a_{0}^{2}

y = (16 a_{0} - 7) x - 8 a_{0}^{2}

y^{3} - (6 x + 8) + 3 x y^{2} y^{^{'}} = 0

4 y^{^{''}} + 4 y^{^{'}} + 1 y = 0, y (2) = - 6, y^{^{'}} (2) = 4

\int_{3}^{6} \frac{x}{x - 3} d x

k \to \infty lim (\frac{1}{k})

\int (sin (x) cos (x))^{2} d x

What is the tangent of 8x^2-7x,\at 9 ?
The tangent of 8x^2-7x,\at 9 is y=(16a_{0}-7)x-8a_{0}^2