What is the tangent of g(x)=x^2+3x ?

The tangent of g(x)=x^2+3x is y=(2a_{0}+3)x-a_{0}^2

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tangent of g(x)=x^2+3x

tangent of

g (x) = x^{2} + 3 x

y = (2 a_{0} + 3) 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} + 3 a_{0})

Find the slope of

g (x) = x^{2} + 3 x : \frac{d g ( x )}{d x} = 2 x + 3

Compute the slope of

g (x) = x^{2} + 3 x

x = a_{0} : m = 2 a_{0} + 3

Find the line with slope m=

2 a_{0} + 3

and passing through

(a_{0}, a_{0}^{2} + 3 a_{0}) : y = (2 a_{0} + 3) x - a_{0}^{2}

y = (2 a_{0} + 3) x - a_{0}^{2}

\frac{d}{d x} (1 + \frac{1}{x ^{4}})

\frac{d}{d x} (cot (x^{2}))

cos (e^{- θ^{2}})

\int (- 9) d x

\int_{- 3}^{3} x^{- \frac{1}{3}} d x

What is the tangent of g(x)=x^2+3x ?
The tangent of g(x)=x^2+3x is y=(2a_{0}+3)x-a_{0}^2