What is the tangent of f(x)=x^2+4x-1 ?

The tangent of f(x)=x^2+4x-1 is y=(2a_{0}+4)x-a_{0}^2-1

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tangent of f(x)=x^2+4x-1

tangent of

f (x) = x^{2} + 4 x - 1

y = (2 a_{0} + 4) x - a_{0}^{2} - 1

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} - 1)

Find the slope of

f (x) = x^{2} + 4 x - 1 : \frac{df ( x )}{d x} = 2 x + 4

Compute the slope of

f (x) = x^{2} + 4 x - 1

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} - 1) : y = (2 a_{0} + 4) x - a_{0}^{2} - 1

y = (2 a_{0} + 4) x - a_{0}^{2} - 1

\int 3 x^{\frac{1}{3}} d x

x - tan (x)

x \to - 4 - lim (\frac{2}{x ^{2} + 3 x - 4})

y^{^{''}} + y^{^{'}} + 20 y = 0

\int_{0}^{1} 6 cos (2 x) d x

What is the tangent of f(x)=x^2+4x-1 ?
The tangent of f(x)=x^2+4x-1 is y=(2a_{0}+4)x-a_{0}^2-1