What is the tangent of f(x)=5-2x^2,\at x=-3,13 ?

The tangent of f(x)=5-2x^2,\at x=-3,13 is y=12x+23

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tangent of f(x)=5-2x^2,\at x=-3,13

tangent of

f (x) = 5 - 2 x^{2}, at x = - 3, 13

y = 12 x + 23

Solution steps

Find the tangent point:

(- 3, - 13)

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}

x = - 3 : m = 12

Find the line with slope m=

12

and passing through

(- 3, - 13) : y = 12 x + 23

y = 12 x + 23

4 x^{\frac{3}{2}}

\int \frac{x}{1 + x} d x

\int \frac{3 ( x + 5 )}{2 x + 15} d x

\int_{0}^{π} 6 sin (x) + sin (6 x) d x

x \to 4 + lim (\frac{6}{x - 4})

What is the tangent of f(x)=5-2x^2,\at x=-3,13 ?
The tangent of f(x)=5-2x^2,\at x=-3,13 is y=12x+23