What is the tangent of 2x^2-8x-7 ?

The tangent of 2x^2-8x-7 is y=(4a_{0}-8)x-2a_{0}^2-7

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tangent of 2x^2-8x-7

tangent of

2 x^{2} - 8 x - 7

y = (4 a_{0} - 8) x - 2 a_{0}^{2} - 7

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

y = 2 x^{2} - 8 x - 7 : \frac{d y}{d x} = 4 x - 8

Compute the slope of

y = 2 x^{2} - 8 x - 7

x = a_{0} : m = 4 a_{0} - 8

Find the line with slope m=

4 a_{0} - 8

and passing through

(a_{0}, 2 a_{0}^{2} - 8 a_{0} - 7) : y = (4 a_{0} - 8) x - 2 a_{0}^{2} - 7

y = (4 a_{0} - 8) x - 2 a_{0}^{2} - 7

\int + cos^{5} (x) d x

\int \frac{5}{8} csc^{2} (\frac{x}{8}) d x

\frac{d}{d x} ((2 x - 4)^{- 4})

\int \frac{1}{( x - 9 ) ( x + 9 )} d x

cos (x) - x sin (x)

What is the tangent of 2x^2-8x-7 ?
The tangent of 2x^2-8x-7 is y=(4a_{0}-8)x-2a_{0}^2-7