Answer is: option3

\( y = 3x - 4 \)

Solution:

We need to determine the equation of the tangent line to the function \( f(x) = x^2 - x \) at the point where \( f'(x) = 3 \).

Differentiate \( f(x) \): \[ f'(x) = \frac{d}{dx} (x^2 - x) = 2x - 1 \] Solve for \( x \) when \( f'(x) = 3 \) \[ 2x - 1 = 3 \] \[ 2x = 4 \] \[ x = 2 \]

Substituting \( x = 2 \) into \( f(x) \):

\[ f(2) = 2^2 - 2 = 4 - 2 = 2 \]

So the point of tangency is \( (2,2) \).

Using the point-slope formula: \[ y - y_1 = m(x - x_1) \]

where \( m = 3 \), \( x_1 = 2 \), and \( y_1 = 2 \):

\[ y - 2 = 3(x - 2) \] Expanding: \[ y = 3x - 6 + 2 \] \[ y = 3x - 4 \]

The correct answer is:

\[ \boxed{y = 3x - 4} \]