What is the tangent of f(x)=19-x^2,\at x=3 ?

The tangent of f(x)=19-x^2,\at x=3 is y=-6x+28

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tangent of f(x)=19-x^2,\at x=3

tangent of

f (x) = 19 - x^{2}, at x = 3

y = - 6 x + 28

Solution steps

Find the tangent point:

(3, 10)

Find the slope of

f (x) = 19 - x^{2} : \frac{df ( x )}{d x} = - 2 x

Compute the slope of

f (x) = 19 - x^{2}

x = 3 : m = - 6

Find the line with slope m=

- 6

and passing through

(3, 10) : y = - 6 x + 28

y = - 6 x + 28

x y^{^{'}} - 2 y^{^{'}} - (x - 2) sin (x) + y = 0

\frac{d}{d x} (\frac{- y}{x})

(- 3, 7), (2, - 5)

f (x) = 1 - 3 x^{2}

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

What is the tangent of f(x)=19-x^2,\at x=3 ?
The tangent of f(x)=19-x^2,\at x=3 is y=-6x+28