What is the tangent of f(x)=x^2-4,(-5,21) ?

The tangent of f(x)=x^2-4,(-5,21) is y=-10x-29

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tangent of f(x)=x^2-4,(-5,21)

tangent of

f (x) = x^{2} - 4, (- 5, 21)

y = - 10 x - 29

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

(- 5, 21) : m = - 10

Find the line with slope m=

- 10

and passing through

(- 5, 21) : y = - 10 x - 29

y = - 10 x - 29

n = 1 \sum \infty \frac{6}{n}

\frac{d}{d x} (2 + 6 x)

\frac{d}{d y} (y^{y})

\frac{d}{d x} (e^{3 x^{2}} + ln (5 x - 4))

\frac{d}{d x} (x - sin^{2} (x) - 4)

What is the tangent of f(x)=x^2-4,(-5,21) ?
The tangent of f(x)=x^2-4,(-5,21) is y=-10x-29