 We can solve rational inequalities from graphs if we keep three ideas in mind. First, the critical values are where the rational expression is zero or fails to exist. And second, the y-values of y equals f of x are the function values. So for example, suppose we have the graphs of y equals f of x and y equals g of x, and let's try and solve f of x divided by g of x is greater than zero. So again, we want to find the critical values. So these will be where our rational expression is either equal to zero or undefined. Now, since it's a quotient, the only times a quotient will be zero is when the numerator is zero. So we want to know when f of x is equal to zero. And on the graph of y equals f of x, f of x is equal to zero whenever y is equal to zero. In other words, at the x-intercepts, and this occurs at x equals negative three or x equals zero. Again, since we have a quotient, the quotient will be undefined whenever the denominator is equal to zero. So we want to know when g of x is equal to zero. And on the graph of y equals g of x, this occurs when y is equal to zero. And that will be here at the x-intercept, which is where x is equal to five. Now in the past, our first step has been to plot those on a number line. So let's go ahead and do that. Conveniently enough, there is a number line right here. And in fact, we've already plotted those points. And we see that there's one, two, three, four regions. Now in this first region, we see that f of x, well, our graph of y equals f of x is below the x-axis. So f of x is less than zero. We also see that the graph of y equals g of x is below the x-axis. And so g of x is also less than zero. And so our quotient f of x divided by g of x, well, that's a negative number, divided by a negative number, which is going to be positive. And so our solution interval should include this first interval from negative infinity up to negative three. Now we have this second interval, which runs between negative three and zero. And so the second region, we see that y equals f of x is above the axis. So f of x is positive. g of x is negative. And so the quotient will be negative. And this second interval is not going to be part of the solution set. Again, it's convenient if we cross it out to indicate that we have actually checked it. This third interval runs from zero up to five. And here we see that f of x and g of x are both negative. And so their quotient will be positive. And we do want to include this region in our solution set. And finally, if we take a look at this last interval from five onward, we see that f of x is negative and g of x is positive. And so our quotient is going to be negative. And so we exclude this last interval. And so our solution interval will just be the first and the third intervals.