What is the tangent of f(x)=x^2+3x+5 ?

The tangent of f(x)=x^2+3x+5 is y=(2a_{0}+3)x-a_{0}^2+5

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tangent of f(x)=x^2+3x+5

tangent of

f (x) = x^{2} + 3 x + 5

y = (2 a_{0} + 3) x - a_{0}^{2} + 5

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} + 5)

Find the slope of

f (x) = x^{2} + 3 x + 5 : \frac{df ( x )}{d x} = 2 x + 3

Compute the slope of

f (x) = x^{2} + 3 x + 5

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} + 5) : y = (2 a_{0} + 3) x - a_{0}^{2} + 5

y = (2 a_{0} + 3) x - a_{0}^{2} + 5

\frac{\partial}{\partial y} (e^{- y})

x \to 1 lim (2 x - 2)

y = \frac{3}{x}

\int \frac{1}{( x ^{2} + 16 ) ^{3}} d x

What is the tangent of f(x)=x^2+3x+5 ?
The tangent of f(x)=x^2+3x+5 is y=(2a_{0}+3)x-a_{0}^2+5