What is the tangent of y=x^2,\at x=-0.5 ?

The tangent of y=x^2,\at x=-0.5 is y=-x-0.25

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tangent of y=x^2,\at x=-0.5

tangent of

y = x^{2}, at x = - 0.5

y = - x - 0.25

Solution steps

Find the tangent point:

(- 0.5, 0.25)

Find the slope of

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

Find the line with slope m=

- 1

and passing through

(- 0.5, 0.25) : y = - x - 0.25

y = - x - 0.25

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

y = (4 x^{6} + 7)^{5}

9 x - 6 x

\frac{\partial}{\partial x} (2 cos (2 x))

\int \frac{5 x + 3}{x ^{2} + 4 x + 10} d x

What is the tangent of y=x^2,\at x=-0.5 ?
The tangent of y=x^2,\at x=-0.5 is y=-x-0.25