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

The tangent of f(x)=5-2x^2,\at x=-2 is y=8x+13

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

tangent of

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

y = 8 x + 13

Solution steps

Find the tangent point:

(- 2, - 3)

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 = - 2 : m = 8

Find the line with slope m=

8

and passing through

(- 2, - 3) : y = 8 x + 13

y = 8 x + 13

\int \frac{2 x + 1}{( - 4 x ^{2} + 4 x + 3 ) ^{\frac{3}{2}}} d x

\frac{d}{d x} (sin (3 x + 2))

\frac{4}{( s + 10 ) ( s ^{2} + 16 ) ^{2}}

y = \frac{x ^{2} + 2 x - 4}{( x + 1 ) ^{2}}

\int \frac{6 x + 5}{x ^{2} + 3 x - 4} d x

What is the tangent of f(x)=5-2x^2,\at x=-2 ?
The tangent of f(x)=5-2x^2,\at x=-2 is y=8x+13