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

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

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

tangent of

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

y = 3 x - 4

Solution steps

Find the tangent point:

(2, 2)

Find the slope of

f (x) = x^{2} - x : \frac{df ( x )}{d x} = 2 x - 1

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

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

y = 3 x - 4

n = 1 \sum \infty \frac{2}{n ( n + 2 )}

\int \frac{sin ( x )}{cos ^{7} ( x )} d x

y^{^{'}} + 8 y - 6 = 0

\int r 1 + r^{2} d r

3 x^{2} e^{x} + e^{x} (x^{3} + 1)

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