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

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

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tangent of x^2-8x

tangent of

x^{2} - 8 x

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

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})

Find the slope of

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

Compute the slope of

y = x^{2} - 8 x

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

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

\int x 6 - x^{2} d x

f (x) = 4 x^{2} + 3 x

y^{^{''}} + 2 y^{^{'}} + 2 y = sin (9 x)

\frac{1}{X ^{2}}

\frac{4 ( - s + 16 )}{s ^{2} + 10 s}