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

The tangent of f(x)=4+3x-2x^2,\at x=2 is y=-5x+12

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

tangent of

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

y = - 5 x + 12

Solution steps

Find the tangent point:

(2, 2)

Find the slope of

f (x) = 4 + 3 x - 2 x^{2} : \frac{df ( x )}{d x} = - 4 x + 3

Compute the slope of

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

x = 2 : m = - 5

Find the line with slope m=

- 5

and passing through

(2, 2) : y = - 5 x + 12

y = - 5 x + 12

y = x (x^{3} + 3)^{5}

f (4 x^{2} - 2)

f (x) = ln (x^{2} - 9)

sin^{2} (+ t)

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

What is the tangent of f(x)=4+3x-2x^2,\at x=2 ?
The tangent of f(x)=4+3x-2x^2,\at x=2 is y=-5x+12