Find the slope of the tangent line to \( y=4 x-x^{2} \) at the point \( (1,3) \) using the definition \( m=\lim _{x \rightarrow a} \frac{f(x)-f(a)}{x-a} \)

To find the slope of the tangent line to the curve \( y = 4x - x^2 \) at the point \( (1, 3) \) using the definition, we start by using the formula \( m = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} \), where \( a = 1 \). Given \( f(x) = 4x - x^2 \), we first compute \( f(1) \): \[ f(1) = 4(1) - (1)^2 = 4 - 1 = 3. \] Now we can substitute this into our formula: \[ m = \lim_{x \to 1} \frac{f(x) - f(1)}{x - 1} = \lim_{x \to 1} \frac{(4x - x^2) - 3}{x - 1}. \] Next, simplify the expression: \[ f(x) - 3 = 4x - x^2 - 3 = -x^2 + 4x - 3. \] Factoring the quadratic expression: \[ -x^2 + 4x - 3 = -(x^2 - 4x + 3) = -(x - 1)(x - 3). \] Substituting back, we have: \[ m = \lim_{x \to 1} \frac{-(x - 1)(x - 3)}{x - 1}. \] For \( x \neq 1 \), we can cancel \( (x - 1) \): \[ m = \lim_{x \to 1} -(x - 3) = - (1 - 3) = -(-2) = 2. \] Thus, the slope of the tangent line to the curve at the point \( (1, 3) \) is \( 2 \).