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

The tangent of f(x)=-x^2+2x+4,\at x=3 is y=-4x+13

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

tangent of

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

y = - 4 x + 13

Solution steps

Find the tangent point:

(3, 1)

Find the slope of

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

Compute the slope of

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

x = 3 : m = - 4

Find the line with slope m=

- 4

and passing through

(3, 1) : y = - 4 x + 13

y = - 4 x + 13

y = - 3 x^{2}

\int \frac{1}{- e ^{2 x}} d x

\frac{d}{d x} (x + \frac{1}{x + 1})

f (x) = \frac{1}{x}, at x = 1

\int e^{- t} cos (t) d t

What is the tangent of f(x)=-x^2+2x+4,\at x=3 ?
The tangent of f(x)=-x^2+2x+4,\at x=3 is y=-4x+13