What is the tangent of y=x^2-1,(-3,8) ?

The tangent of y=x^2-1,(-3,8) is y=-6x-10

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tangent of y=x^2-1,(-3,8)

tangent of

y = x^{2} - 1, (- 3, 8)

y = - 6 x - 10

Solution steps

Find the slope of

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

Compute the slope of

y = x^{2} - 1

(- 3, 8) : m = - 6

Find the line with slope m=

- 6

and passing through

(- 3, 8) : y = - 6 x - 10

y = - 6 x - 10

\frac{\partial}{\partial x} (6 x - 5 y)

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

\int \frac{289 - x ^{2}}{x} d x

x \to 0 lim ((1 + x)^{\frac{5}{x}})

x^{2} \frac{d y}{d x} - 2 x y = 6 y^{4}, y (1) = \frac{1}{2}

What is the tangent of y=x^2-1,(-3,8) ?
The tangent of y=x^2-1,(-3,8) is y=-6x-10