Answer & step-by-step explanation:
is simply calculating by replacing a in lieu of x

Same difference for
, just be careful with products and squares:

In the last step i just reorganized terms so that everything containing h was at the end.
Finally, for the incremental ratio, writing it as 1/h for the sake of font sizes:
![\frac{f(a+h)-f(a)}h = \frac1h[(3-3a+5a^2-3h+10ah+5h^2)-(3-3a+5a^2)]=\\\frac1h(-3h+10ah+5h^2) = 10a-3 +5h](https://img.qammunity.org/2023/formulas/mathematics/college/wj6owb1hhe6gpxpb8sn71z4yl3ow09sk2a.png)