What is the tangent of f(x)=675-4.9x^2 ?

The tangent of f(x)=675-4.9x^2 is y=-9.8a_{0}x+(6750+49a_{0}^2)/(10)

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tangent of f(x)=675-4.9x^2

tangent of

f (x) = 675 - 4.9 x^{2}

y = - 9.8 a_{0} x + \frac{6750 + 49 a _{0}^{2}}{10}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 675 - 4.9 a_{0}^{2})

Find the slope of

f (x) = 675 - 4.9 x^{2} : \frac{df ( x )}{d x} = - 9.8 x

Compute the slope of

f (x) = 675 - 4.9 x^{2}

x = a_{0} : m = - 9.8 a_{0}

Find the line with slope m=

- 9.8 a_{0}

and passing through

(a_{0}, 675 - 4.9 a_{0}^{2}) : y = - 9.8 a_{0} x + \frac{6750 + 49 a _{0}^{2}}{10}

y = - 9.8 a_{0} x + \frac{6750 + 49 a _{0}^{2}}{10}

\frac{d}{d x} (7^{x^{3}})

2 y^{^{''}} - 3 y^{^{'}} + 4 y = 0

\frac{\partial}{\partial z} (- x^{3} yz e^{z})

\frac{d}{d x} (1000 e^{- \frac{x}{10}})

\int x^{2} (8 x^{3} + 7)^{2} d x

What is the tangent of f(x)=675-4.9x^2 ?
The tangent of f(x)=675-4.9x^2 is y=-9.8a_{0}x+(6750+49a_{0}^2)/(10)