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

The tangent of f(x)=-x^2+2x+4,\at x=4 is y=-6x+20

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

tangent of

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

y = - 6 x + 20

Solution steps

Find the tangent point:

(4, - 4)

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 = 4 : m = - 6

Find the line with slope m=

- 6

and passing through

(4, - 4) : y = - 6 x + 20

y = - 6 x + 20

\int_{0}^{1} \frac{1}{e ^{3 x}} d x

\frac{d}{d x} (e^{- a x^{2}})

n = 0 \sum \infty \frac{e ^{n}}{π ^{n}}

\frac{d}{d x} ((2 x - 8)^{4} (x^{2} + x + 1)^{5})

\frac{d}{d x} (3 (4 e^{2 x} x + 4 e^{2 x}))

What is the tangent of f(x)=-x^2+2x+4,\at x=4 ?
The tangent of f(x)=-x^2+2x+4,\at x=4 is y=-6x+20