What is the tangent of y=x^2+3x ?

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

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tangent of y=x^2+3x

tangent of

y = 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

y = x^{2} + 3 x : \frac{d y}{d x} = 2 x + 3

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : 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}

e^{3 x}, 1

arctan (\frac{1}{x})

f (x) = \frac{x ^{3} - x ^{2} - 1}{x ^{2}}

x \to - 2 lim (2 + x)

x \to 0 lim (\frac{x - 1}{x})

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