1. Find and simplify the difference quotient for for \( f(x)=\frac{-5 x^{2}}{2}-3 x \). Show all work.

To find the difference quotient for \( f(x) = \frac{-5 x^{2}}{2} - 3 x \), we start by calculating \( f(x+h) \): \[ f(x+h) = \frac{-5 (x+h)^{2}}{2} - 3 (x+h). \] Expanding \( (x+h)^{2} \): \[ (x+h)^{2} = x^{2} + 2xh + h^{2}, \] so \[ f(x+h) = \frac{-5 (x^{2} + 2xh + h^{2})}{2} - 3x - 3h. \] This simplifies to: \[ f(x+h) = \frac{-5x^{2}}{2} - 5xh - \frac{5h^{2}}{2} - 3x - 3h. \] Now put \( f(x+h) \) and \( f(x) \) into the difference quotient: \[ \frac{f(x+h) - f(x)}{h} = \frac{\left(\frac{-5x^{2}}{2} - 5xh - \frac{5h^{2}}{2} - 3x - 3h\right) - \left(\frac{-5x^{2}}{2} - 3x\right)}{h}. \] Cancelling \( \frac{-5x^{2}}{2} \) and \( -3x \): \[ = \frac{-5xh - \frac{5h^{2}}{2} - 3h}{h}. \] Factoring \( h \) out of the numerator gives: \[ = \frac{h\left(-5x - \frac{5h}{2} -3\right)}{h}. \] Since \( h \neq 0 \), we can cancel \( h \): \[ = -5x - \frac{5h}{2} - 3. \] Thus, the simplified difference quotient is: \[ -5x - 3 - \frac{5h}{2}. \]