What is the tangent of x^2+4x-5 ?

The tangent of x^2+4x-5 is y=(2a_{0}+4)x-a_{0}^2-5

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tangent of x^2+4x-5

tangent of

x^{2} + 4 x - 5

y = (2 a_{0} + 4) x - a_{0}^{2} - 5

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{2} + 4 a_{0} - 5)

Find the slope of

y = x^{2} + 4 x - 5 : \frac{d y}{d x} = 2 x + 4

Compute the slope of

y = x^{2} + 4 x - 5

x = a_{0} : m = 2 a_{0} + 4

Find the line with slope m=

2 a_{0} + 4

and passing through

(a_{0}, a_{0}^{2} + 4 a_{0} - 5) : y = (2 a_{0} + 4) x - a_{0}^{2} - 5

y = (2 a_{0} + 4) x - a_{0}^{2} - 5

\frac{d y}{d x} = \frac{( x - 1 )}{( y - 1 )}

\int_{0}^{0.1} x arctan (4 x) d x

\frac{d y}{d t} = \frac{8 t y}{t ^{2} + 1}

y^{^{'}} = 1 - t + y^{2} - t y^{2}

\frac{d}{d x} (- 7 sin (x))

What is the tangent of x^2+4x-5 ?
The tangent of x^2+4x-5 is y=(2a_{0}+4)x-a_{0}^2-5