For the function \( f(x)=5 x^{2}+5 x-5 \), evaluate and fully simplify each of the following. \( f(x+h)=\square \) \( \frac{f(x+h)-f(x)}{h}=\square \)

To find \( f(x+h) \) for the function \( f(x) = 5x^2 + 5x - 5 \), we substitute \( x+h \) into the function: \[ f(x+h) = 5(x+h)^2 + 5(x+h) - 5 \] Expanding \( (x+h)^2 \) gives \( x^2 + 2xh + h^2 \), so: \[ f(x+h) = 5(x^2 + 2xh + h^2) + 5x + 5h - 5 \] \[ = 5x^2 + 10xh + 5h^2 + 5x + 5h - 5 \] Combining like terms results in: \[ f(x+h) = 5x^2 + (10xh + 5x + 5h - 5). \] Next, we need to calculate \( \frac{f(x+h) - f(x)}{h} \): 1. We know \( f(x) = 5x^2 + 5x - 5 \). 2. Substitute \( f(x+h) \): \[ \frac{f(x+h) - f(x)}{h} = \frac{(5x^2 + 10xh + 5h^2 + 5x + 5h - 5) - (5x^2 + 5x - 5)}{h} \] 3. Simplifying within the brackets: \[ = \frac{10xh + 5h^2 + 5h}{h} \] 4. Factoring out \( h \) in the numerator yields: \[ = \frac{h(10x + 5h + 5)}{h} \] 5. Cancel the \( h \): \[ = 10x + 5h + 5 \] Finally, the evaluations are: \( f(x+h) = 5x^2 + 10xh + 5h^2 + 5x + 5h - 5 \) \( \frac{f(x+h) - f(x)}{h} = 10x + 5h + 5 \)