What is the tangent of f(x)=2-5x^2,(-2,-18) ?

The tangent of f(x)=2-5x^2,(-2,-18) is y=20x+22

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tangent of f(x)=2-5x^2,(-2,-18)

tangent of

f (x) = 2 - 5 x^{2}, (- 2, - 18)

y = 20 x + 22

Solution steps

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 20

Find the line with slope m=

20

and passing through

(- 2, - 18) : y = 20 x + 22

y = 20 x + 22

\frac{x - 2}{( x + 2 ) ( x - 3 )}

y^{^{'}} = \frac{2 y}{x}, y (\frac{1}{2}) = 5

x \to 0 lim (\frac{( x ^{3} )}{x})

\int_{- 1}^{3} x d x

\int (2 x^{3} + 5 x^{2} - 2 x + 1) d x

What is the tangent of f(x)=2-5x^2,(-2,-18) ?
The tangent of f(x)=2-5x^2,(-2,-18) is y=20x+22