What is the tangent of f(x)=-3x^2+8x+1,\at x=-2 ?

The tangent of f(x)=-3x^2+8x+1,\at x=-2 is y=20x+13

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tangent of f(x)=-3x^2+8x+1,\at x=-2

tangent of

f (x) = - 3 x^{2} + 8 x + 1, at x = - 2

y = 20 x + 13

Solution steps

Find the tangent point:

(- 2, - 27)

Find the slope of

f (x) = - 3 x^{2} + 8 x + 1 : \frac{df ( x )}{d x} = - 6 x + 8

Compute the slope of

f (x) = - 3 x^{2} + 8 x + 1

x = - 2 : m = 20

Find the line with slope m=

20

and passing through

(- 2, - 27) : y = 20 x + 13

y = 20 x + 13

3^{- x}

x \to 6 lim (- x^{2} + 4 x - 3)

\int csc (x) sec^{2} (x) d x

\int \frac{x ^{2} + 1}{x ^{2} - 5 x + 6} d x

\frac{\partial}{\partial x} (5 e^{2 x})

What is the tangent of f(x)=-3x^2+8x+1,\at x=-2 ?
The tangent of f(x)=-3x^2+8x+1,\at x=-2 is y=20x+13