Rate = k’[CV], where k’ = k[NaOH]y

Question

Look at the graphs for Run 1 and Run 2 that pertain to the order of the reaction with respect to [CV]. The slope of the graphs is equal to k’, which is itself equal to k[NaOH]^y. In the first run, [NaOH] = 0.010 M after dilution, and in the second run, [NaOH] = 0.020 M after dilution. Use this information
to find the value for “y”.

I have the graphs but the wording for this problem on setting it up and finding initial rate is confusing.