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

The tangent of f(x)=4x^2+3x is y=(8a_{0}+3)x-4a_{0}^2

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

tangent of

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

y = (8 a_{0} + 3) x - 4 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

f (x) = 4 x^{2} + 3 x : \frac{df ( x )}{d x} = 8 x + 3

Compute the slope of

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

x = a_{0} : m = 8 a_{0} + 3

Find the line with slope m=

8 a_{0} + 3

and passing through

(a_{0}, 4 a_{0}^{2} + 3 a_{0}) : y = (8 a_{0} + 3) x - 4 a_{0}^{2}

y = (8 a_{0} + 3) x - 4 a_{0}^{2}

\int x^{3} x^{2} + 25 d x

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\frac{\partial}{\partial x} (2 (x e^{\frac{x}{2} + y} + 1))

\frac{d}{d x} (2 e^{x} + 3 x - ln (x))

\int - \frac{1}{2} sin (2 x) d x

What is the tangent of f(x)=4x^2+3x ?
The tangent of f(x)=4x^2+3x is y=(8a_{0}+3)x-4a_{0}^2