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

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

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

tangent of

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

y = (2 a_{0} + 4) 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} + 4 a_{0})

Find the slope of

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

Compute the slope of

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

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

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

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