What is the tangent of f(x)=5x^2+4 ?

The tangent of f(x)=5x^2+4 is y=10a_{0}x-5a_{0}^2+4

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tangent of f(x)=5x^2+4

tangent of

f (x) = 5 x^{2} + 4

y = 10 a_{0} x - 5 a_{0}^{2} + 4

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 5 a_{0}^{2} + 4)

Find the slope of

f (x) = 5 x^{2} + 4 : \frac{df ( x )}{d x} = 10 x

Compute the slope of

f (x) = 5 x^{2} + 4

x = a_{0} : m = 10 a_{0}

Find the line with slope m=

10 a_{0}

and passing through

(a_{0}, 5 a_{0}^{2} + 4) : y = 10 a_{0} x - 5 a_{0}^{2} + 4

y = 10 a_{0} x - 5 a_{0}^{2} + 4

\frac{d}{d x} (\frac{- 2}{2 x + 5})

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

\frac{d y}{d t} + e^{t} y = - 24 e^{t}

\int 3^{x + 2} d x

(sin (x) + cos (x))^{^{'}}

What is the tangent of f(x)=5x^2+4 ?
The tangent of f(x)=5x^2+4 is y=10a_{0}x-5a_{0}^2+4