Find the difference quotient of \( f \), that is, find \( \frac{f(x+h)-f(x)}{h}, h \neq 0 \), for the following function. Be sure to simplify. \( f(x)=x^{2}-4 x+5 \) \( \frac{f(x+h)-f(x)}{h}=\square \)

To find the difference quotient \( \frac{f(x+h)-f(x)}{h} \) for the function \( f(x) = x^2 - 4x + 5 \), we first compute \( f(x+h) \): \[ f(x+h) = (x+h)^2 - 4(x+h) + 5 \] Expanding this, we get: \[ f(x+h) = (x^2 + 2xh + h^2) - (4x + 4h) + 5 = x^2 + 2xh + h^2 - 4x - 4h + 5 \] Now, we can substitute into the difference quotient: \[ f(x+h) - f(x) = (x^2 + 2xh + h^2 - 4x - 4h + 5) - (x^2 - 4x + 5) \] Simplifying this expression, we notice the \( x^2 \), \( -4x \), and \( +5 \) will cancel out: \[ f(x+h) - f(x) = 2xh + h^2 - 4h \] Now we plug this result into the difference quotient: \[ \frac{f(x+h) - f(x)}{h} = \frac{2xh + h^2 - 4h}{h} \] We can simplify by dividing each term by \( h \) (since \( h \neq 0 \)): \[ \frac{f(x+h) - f(x)}{h} = 2x + h - 4 \] Thus, the simplified difference quotient is: \[ \frac{f(x+h)-f(x)}{h} = 2x + h - 4 \]