What is the tangent of f(x)=625-4.9x^2,\at x=4 ?

The tangent of f(x)=625-4.9x^2,\at x=4 is y=-39.2x+703.4

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tangent of f(x)=625-4.9x^2,\at x=4

tangent of

f (x) = 625 - 4.9 x^{2}, at x = 4

y = - 39.2 x + 703.4

Solution steps

Find the tangent point:

(4, 546.6)

Find the slope of

f (x) = 625 - 4.9 x^{2} : \frac{df ( x )}{d x} = - 9.8 x

Compute the slope of

f (x) = 625 - 4.9 x^{2}

x = 4 : m = - 39.2

Find the line with slope m=

- 39.2

and passing through

(4, 546.6) : y = - 39.2 x + 703.4

y = - 39.2 x + 703.4

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

(0, 0), (8, 6)

n = 0 \sum \infty (1.67)^{n}

\frac{d y}{d x}, y^{9} = x^{8}

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

What is the tangent of f(x)=625-4.9x^2,\at x=4 ?
The tangent of f(x)=625-4.9x^2,\at x=4 is y=-39.2x+703.4