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

The tangent of f(x)=x^2-4,(-3,5) is y=-6x-13

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tangent of f(x)=x^2-4,(-3,5)

tangent of

f (x) = x^{2} - 4, (- 3, 5)

y = - 6 x - 13

Solution steps

Find the slope of

f (x) = x^{2} - 4 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = x^{2} - 4

(- 3, 5) : m = - 6

Find the line with slope m=

- 6

and passing through

(- 3, 5) : y = - 6 x - 13

y = - 6 x - 13

\frac{d y}{d t} = 1.5 y, y (0) = 100

\int_{0}^{\infty} t e^{- b t} d t

f (x) = - 8 x^{4} + 11 x^{2} - 3, (1, 0)

\int \frac{2}{x ^{2} + 3 x} d x

\int \frac{( sin ( x ) )}{cos ( x )} d x

What is the tangent of f(x)=x^2-4,(-3,5) ?
The tangent of f(x)=x^2-4,(-3,5) is y=-6x-13