What is the tangent of f(x)=8x^2+3x ?

The tangent of f(x)=8x^2+3x is y=(16a_{0}+3)x-8a_{0}^2

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tangent of f(x)=8x^2+3x

tangent of

f (x) = 8 x^{2} + 3 x

y = (16 a_{0} + 3) x - 8 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 8 a_{0}^{2} + 3 a_{0})

Find the slope of

f (x) = 8 x^{2} + 3 x : \frac{df ( x )}{d x} = 16 x + 3

Compute the slope of

f (x) = 8 x^{2} + 3 x

x = a_{0} : m = 16 a_{0} + 3

Find the line with slope m=

16 a_{0} + 3

and passing through

(a_{0}, 8 a_{0}^{2} + 3 a_{0}) : y = (16 a_{0} + 3) x - 8 a_{0}^{2}

y = (16 a_{0} + 3) x - 8 a_{0}^{2}

\frac{\partial}{\partial x} (2 y - \frac{1}{x} + cos (3 x))

\int x^{3} x^{4} + 9 d x

\frac{d}{d x} (2 sin (x) + 3 cos (x))

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

x \to 1 lim (\frac{x + 4}{x + 2})

What is the tangent of f(x)=8x^2+3x ?
The tangent of f(x)=8x^2+3x is y=(16a_{0}+3)x-8a_{0}^2