What is the tangent of f(x)=2x^2+4x+3 ?

The tangent of f(x)=2x^2+4x+3 is y=(4a_{0}+4)x-2a_{0}^2+3

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tangent of f(x)=2x^2+4x+3

tangent of

f (x) = 2 x^{2} + 4 x + 3

y = (4 a_{0} + 4) x - 2 a_{0}^{2} + 3

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2 a_{0}^{2} + 4 a_{0} + 3)

Find the slope of

f (x) = 2 x^{2} + 4 x + 3 : \frac{df ( x )}{d x} = 4 x + 4

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 = 4 a_{0} + 4

Find the line with slope m=

4 a_{0} + 4

and passing through

(a_{0}, 2 a_{0}^{2} + 4 a_{0} + 3) : y = (4 a_{0} + 4) x - 2 a_{0}^{2} + 3

y = (4 a_{0} + 4) x - 2 a_{0}^{2} + 3

\frac{d y}{d x} = \frac{2 y}{x}, x > 0

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

\frac{\partial}{\partial x} (ln (x + y^{3} + x y^{2}))

\frac{\partial}{\partial z} (x^{2} y z^{2})

\frac{d}{d x} (1 - \frac{3}{x + 7})

What is the tangent of f(x)=2x^2+4x+3 ?
The tangent of f(x)=2x^2+4x+3 is y=(4a_{0}+4)x-2a_{0}^2+3