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

The tangent of x^2-5x,\at x=4 is y=3x-16

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tangent of x^2-5x,\at x=4

tangent of

x^{2} - 5 x, at x = 4

y = 3 x - 16

Solution steps

Find the tangent point:

(4, - 4)

Find the slope of

y = x^{2} - 5 x : \frac{d y}{d x} = 2 x - 5

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 = 3

Find the line with slope m=

3

and passing through

(4, - 4) : y = 3 x - 16

y = 3 x - 16

\frac{d}{d x} (arctan (x^{3}))

d er i v a t i v e e^{- s t} d t

\frac{d}{d x} (\frac{x ^{2} - 2 x + 2}{x - 1})

\int 1 + 4 y^{2} d y

d er i v a t i v e \frac{ln ( x )}{x ^{3}}

What is the tangent of x^2-5x,\at x=4 ?
The tangent of x^2-5x,\at x=4 is y=3x-16