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

The tangent of f(x)=-x^2+4x,\at x=3 is y=-2x+9

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

tangent of

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

y = - 2 x + 9

Solution steps

Find the tangent point:

(3, 3)

Find the slope of

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

Compute the slope of

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

x = 3 : m = - 2

Find the line with slope m=

- 2

and passing through

(3, 3) : y = - 2 x + 9

y = - 2 x + 9

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

\frac{d X}{d X}

5 e^{- 3 (t - 2)}

y^{^{''}} - 2 y^{^{'}} + y = 2 t - 1 - (6 t e^{t} + 6 e^{t})

1 + T^{2}

What is the tangent of f(x)=-x^2+4x,\at x=3 ?
The tangent of f(x)=-x^2+4x,\at x=3 is y=-2x+9