Try this:

=(G9-MIN($G$9:$G:$12))/(MAX($G$9:$G$12)-MIN($G$9:$G$12))

If you want to normalize your data, you can do so as you suggest and simply calculate the following:

$$z_i=\frac{x_i-\min(x)}{\max(x)-\min(x)}$$

where $x=(x_1,...,x_n)$ and $z_i$ is now your $i^{th}$ normalized data. As a proof of concept (although you did not ask for it) here is some R code and accompanying graph to illustrate this point:

# Example Data
x = sample(-100:100, 50)

#Normalized Data
normalized = (x-min(x))/(max(x)-min(x))

# Histogram of example data and normalized data
par(mfrow=c(1,2))
hist(x,          breaks=10, xlab="Data",            col="lightblue", main="")
hist(normalized, breaks=10, xlab="Normalized Data", col="lightblue", main="")