What is the tangent of y=x^2-8x,\at x=-8 ?

The tangent of y=x^2-8x,\at x=-8 is y=-24x-64

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tangent of y=x^2-8x,\at x=-8

tangent of

y = x^{2} - 8 x, at x = - 8

y = - 24 x - 64

Solution steps

Find the tangent point:

(- 8, 128)

Find the slope of

y = x^{2} - 8 x : \frac{d y}{d x} = 2 x - 8

Compute the slope of

y = x^{2} - 8 x

x = - 8 : m = - 24

Find the line with slope m=

- 24

and passing through

(- 8, 128) : y = - 24 x - 64

y = - 24 x - 64

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What is the tangent of y=x^2-8x,\at x=-8 ?
The tangent of y=x^2-8x,\at x=-8 is y=-24x-64