What is the tangent of f(x)=5-2x^2,(-5,-45) ?

The tangent of f(x)=5-2x^2,(-5,-45) is y=20x+55

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tangent of f(x)=5-2x^2,(-5,-45)

tangent of

f (x) = 5 - 2 x^{2}, (- 5, - 45)

y = 20 x + 55

Solution steps

Find the slope of

f (x) = 5 - 2 x^{2} : \frac{df ( x )}{d x} = - 4 x

Compute the slope of

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

(- 5, - 45) : m = 20

Find the line with slope m=

20

and passing through

(- 5, - 45) : y = 20 x + 55

y = 20 x + 55

\int_{0}^{2} (1 + x)^{\frac{1}{3}} \cdot 2 d x

x y^{'} + (1 + x \cdot cot (x)) y = 0

\frac{d}{d x} ((x - 1)^{2} (x + 2))

y = 3 - x^{2}, y = 11 - 6 x

\int_{0}^{4} \frac{x}{1 + 2 x} d x

What is the tangent of f(x)=5-2x^2,(-5,-45) ?
The tangent of f(x)=5-2x^2,(-5,-45) is y=20x+55