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

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

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tangent of x^2+8x-7

tangent of

x^{2} + 8 x - 7

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

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} - 7)

Find the slope of

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

Compute the slope of

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

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

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

\frac{9 x}{2 ^{x}}

y = 4 x^{2} + 2

x \to - 2 lim (\frac{x ^{2} - 3}{x + 2})

\frac{\partial}{\partial y} (e^{x^{2} - y})

\int 3 x + \frac{10}{x ^{6}} d x

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