What is the tangent of f(x)=2x^2-1,\at x=-1 ?

The tangent of f(x)=2x^2-1,\at x=-1 is y=-4x-3

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tangent of f(x)=2x^2-1,\at x=-1

tangent of

f (x) = 2 x^{2} - 1, at x = - 1

y = - 4 x - 3

Solution steps

Find the tangent point:

(- 1, 1)

Find the slope of

f (x) = 2 x^{2} - 1 : \frac{df ( x )}{d x} = 4 x

Compute the slope of

f (x) = 2 x^{2} - 1

x = - 1 : m = - 4

Find the line with slope m=

- 4

and passing through

(- 1, 1) : y = - 4 x - 3

y = - 4 x - 3

\frac{d y}{d x} = x^{2} y - y + x^{2} - 1

\int 3 x^{4} d x

\frac{1}{s ^{2} ( s ^{2} + 1 )}

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

\int x^{5} sin (3 + x^{\frac{7}{2}}) d x

What is the tangent of f(x)=2x^2-1,\at x=-1 ?
The tangent of f(x)=2x^2-1,\at x=-1 is y=-4x-3