What is the tangent of f(x)=38-x^2,(-6,2) ?

The tangent of f(x)=38-x^2,(-6,2) is y=12x+74

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tangent of f(x)=38-x^2,(-6,2)

tangent of

f (x) = 38 - x^{2}, (- 6, 2)

y = 12 x + 74

Solution steps

Find the slope of

f (x) = 38 - x^{2} : \frac{df ( x )}{d x} = - 2 x

Compute the slope of

f (x) = 38 - x^{2}

(- 6, 2) : m = 12

Find the line with slope m=

12

and passing through

(- 6, 2) : y = 12 x + 74

y = 12 x + 74

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What is the tangent of f(x)=38-x^2,(-6,2) ?
The tangent of f(x)=38-x^2,(-6,2) is y=12x+74