What is the tangent of f(x)=6x^2-3,\at x=-9 ?

The tangent of f(x)=6x^2-3,\at x=-9 is y=-108x-489

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tangent of f(x)=6x^2-3,\at x=-9

tangent of

f (x) = 6 x^{2} - 3, at x = - 9

y = - 108 x - 489

Solution steps

Find the tangent point:

(- 9, 483)

Find the slope of

f (x) = 6 x^{2} - 3 : \frac{df ( x )}{d x} = 12 x

Compute the slope of

f (x) = 6 x^{2} - 3

x = - 9 : m = - 108

Find the line with slope m=

- 108

and passing through

(- 9, 483) : y = - 108 x - 489

y = - 108 x - 489

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What is the tangent of f(x)=6x^2-3,\at x=-9 ?
The tangent of f(x)=6x^2-3,\at x=-9 is y=-108x-489