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

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

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

tangent of

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

y = - 4 x + 13

Solution steps

Find the tangent point:

(3, 1)

Find the slope of

y = - x^{2} + 2 x + 4 : \frac{d y}{d x} = - 2 x + 2

Compute the slope of

y = - 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

f (x) = \frac{4 x}{x ^{2} + 1}, at x = 1

\int \frac{1}{x x ^{2} - 5} d x

\int (x + \frac{2}{x}) (x - \frac{2}{x}) d x

\int - 14 x d x

\int \frac{1}{3} e^{3 x + 1} d x

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