What is the tangent of f(x)=2-5x^2,(3,-43) ?

The tangent of f(x)=2-5x^2,(3,-43) is y=-30x+47

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tangent of f(x)=2-5x^2,(3,-43)

tangent of

f (x) = 2 - 5 x^{2}, (3, - 43)

y = - 30 x + 47

Solution steps

Find the slope of

f (x) = 2 - 5 x^{2} : \frac{df ( x )}{d x} = - 10 x

Compute the slope of

f (x) = 2 - 5 x^{2}

(3, - 43) : m = - 30

Find the line with slope m=

- 30

and passing through

(3, - 43) : y = - 30 x + 47

y = - 30 x + 47

(csc (t) + cot (t))^{- 1}

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

x \to 0 lim (\frac{( tan ( x ) )}{x})

\int_{0}^{1} x^{2} - 1 d x

\int (3 x^{4} + x^{5}) d x

What is the tangent of f(x)=2-5x^2,(3,-43) ?
The tangent of f(x)=2-5x^2,(3,-43) is y=-30x+47