What is the tangent of f(x)=9-3x+2x^2,\at x=-6 ?

The tangent of f(x)=9-3x+2x^2,\at x=-6 is y=-27x-63

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tangent of f(x)=9-3x+2x^2,\at x=-6

tangent of

f (x) = 9 - 3 x + 2 x^{2}, at x = - 6

y = - 27 x - 63

Solution steps

Find the tangent point:

(- 6, 99)

Find the slope of

f (x) = 9 - 3 x + 2 x^{2} : \frac{df ( x )}{d x} = 4 x - 3

Compute the slope of

f (x) = 9 - 3 x + 2 x^{2}

x = - 6 : m = - 27

Find the line with slope m=

- 27

and passing through

(- 6, 99) : y = - 27 x - 63

y = - 27 x - 63

\frac{d}{d x} (2 x - 3 x^{\frac{2}{3}})

\int 25 e^{- 2 x} d x

x \to 3 - lim (x - 2)

x \to \frac{π}{2} - lim (cot (2 x))

\frac{d y}{d x} = x^{3} y

What is the tangent of f(x)=9-3x+2x^2,\at x=-6 ?
The tangent of f(x)=9-3x+2x^2,\at x=-6 is y=-27x-63