What is the tangent of f(x)=3-x^2,\at x=4 ?

The tangent of f(x)=3-x^2,\at x=4 is y=-8x+19

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tangent of f(x)=3-x^2,\at x=4

tangent of

f (x) = 3 - x^{2}, at x = 4

y = - 8 x + 19

Solution steps

Find the tangent point:

(4, - 13)

Find the slope of

f (x) = 3 - x^{2} : \frac{df ( x )}{d x} = - 2 x

Compute the slope of

f (x) = 3 - x^{2}

x = 4 : m = - 8

Find the line with slope m=

- 8

and passing through

(4, - 13) : y = - 8 x + 19

y = - 8 x + 19

y^{^{'}} = 5 - 2 y

y^{^{'}} = cos^{2} (y)

y = \frac{x ^{2}}{x + 1}, (2, \frac{4}{3})

\frac{d y}{d x} + 6 x y^{5} = 0

(7 x + 11 y) d x + (11 x + 5 y) d y = 0, y (0) = 2

What is the tangent of f(x)=3-x^2,\at x=4 ?
The tangent of f(x)=3-x^2,\at x=4 is y=-8x+19