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

The tangent of f(x)=x^2-4x,\at x=3 is y=2x-9

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

tangent of

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

y = 2 x - 9

Solution steps

Find the tangent point:

(3, - 3)

Find the slope of

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

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 = 2

Find the line with slope m=

2

and passing through

(3, - 3) : y = 2 x - 9

y = 2 x - 9

\int (2 x + 1)^{3} d x

n = 1 \sum \infty \frac{n ^{2}}{5 n}

\frac{d}{d t} (\frac{1}{1 + t})

f (x) = (x + 2)^{\frac{1}{2}}

n \to \infty lim (\frac{( 5 n )}{5 n + 7})

What is the tangent of f(x)=x^2-4x,\at x=3 ?
The tangent of f(x)=x^2-4x,\at x=3 is y=2x-9