What is the tangent of f(x)=8x-3x^2,(-3,-51) ?

The tangent of f(x)=8x-3x^2,(-3,-51) is y=26x+27

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tangent of f(x)=8x-3x^2,(-3,-51)

tangent of

f (x) = 8 x - 3 x^{2}, (- 3, - 51)

y = 26 x + 27

Solution steps

Find the slope of

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

Compute the slope of

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

(- 3, - 51) : m = 26

Find the line with slope m=

26

and passing through

(- 3, - 51) : y = 26 x + 27

y = 26 x + 27

e^{3}

f (x) = - 3 x

\frac{d}{d x} (4 - x^{2} - y^{2})

lo g_{10} (x + 5)

x (\frac{d y}{d x}) + 3 y = \frac{1}{x}

What is the tangent of f(x)=8x-3x^2,(-3,-51) ?
The tangent of f(x)=8x-3x^2,(-3,-51) is y=26x+27