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

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

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

tangent of

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

y = - 2 x - 1

Solution steps

Find the tangent point:

(- 1, 1)

Find the slope of

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

Find the line with slope m=

- 2

and passing through

(- 1, 1) : y = - 2 x - 1

y = - 2 x - 1

\int \frac{1}{z} d z

\frac{d x}{d t} + 5 t x^{3} + \frac{x}{t} = 0

x y y^{^{'}} = x^{2} + 1

\int 4 x ln (x) d x

x^{2} - x y + y^{2} = 3, (1, 2)

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