What is the tangent of y=x^2+5x+9 ?

The tangent of y=x^2+5x+9 is y=(2a_{0}+5)x-a_{0}^2+9

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tangent of y=x^2+5x+9

tangent of

y = x^{2} + 5 x + 9

y = (2 a_{0} + 5) x - a_{0}^{2} + 9

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{2} + 5 a_{0} + 9)

Find the slope of

y = x^{2} + 5 x + 9 : \frac{d y}{d x} = 2 x + 5

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} + 5

Find the line with slope m=

2 a_{0} + 5

and passing through

(a_{0}, a_{0}^{2} + 5 a_{0} + 9) : y = (2 a_{0} + 5) x - a_{0}^{2} + 9

y = (2 a_{0} + 5) x - a_{0}^{2} + 9

x \to 0 lim (\frac{5}{2 x})

n = 2 \sum \infty \frac{1}{( 2 n + 7 ) ^{3}}

e^{3 x^{3}}

\int_{1}^{2} (x^{5} + x^{- 5}) d x

\int x^{3} (2 + x^{4})^{5} d x

What is the tangent of y=x^2+5x+9 ?
The tangent of y=x^2+5x+9 is y=(2a_{0}+5)x-a_{0}^2+9