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

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

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

tangent of

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

y = - 10 x - 29

Solution steps

Find the tangent point:

(- 5, 21)

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 10

Find the line with slope m=

- 10

and passing through

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

y = - 10 x - 29

\int \frac{x}{3 x ^{2} + 1} d x

t a y l or cos (x), \frac{π}{4}

\int sin^{2} (x co) s^{4} x d x

x \to 0 + lim (sin (x) ln (3 x))

d er i v a t i v e x^{2} - x + 4

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