What is the tangent of y=x^2-1,(-2,3) ?

The tangent of y=x^2-1,(-2,3) is y=-4x-5

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tangent of y=x^2-1,(-2,3)

tangent of

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

y = - 4 x - 5

Solution steps

Find the slope of

y = x^{2} - 1 : \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 = - 4

Find the line with slope m=

- 4

and passing through

(- 2, 3) : y = - 4 x - 5

y = - 4 x - 5

f (x) = 4 x^{2} - 4 x + 1, at x = 4

\frac{\partial}{\partial x} (y \cdot e^{x})

f (x) = \frac{x}{x + 3}

n = 1 \sum \infty \frac{( - 1 ) ^{n + 1} 2 n}{n + 1}

\int_{- \frac{π}{4}}^{0} 6 sec^{3} (x) d x

What is the tangent of y=x^2-1,(-2,3) ?
The tangent of y=x^2-1,(-2,3) is y=-4x-5