Let $x_a$ denote amount of feed A and $x_b$ denote amount of feed B.
Minimize $10x_a+12x_b$
subject to
$4x_a+2x_b \ge 12$
$4x_a+8x_b \ge 24$
$8x_b \ge 8$
Have I got the numbers right?
Also to solve it graphically I would get 3 lines that intersect at most 3 times and just test each vertex point, yes?
You are on the right track. You can test each vertex by calculation. But you can evaluate graphically the optimal point as well. I´ve solved the constraints for $x_b$. Therefore the y-axis is denoted as $x_b$. If I understand you right you are able to draw the constraints into the coordinate system. It follows that the feasible region is the green one.
Let´s denote $z$ as the value of the objective function. Then we have the equality
$z=10x_a+12x_b$
This equality can be solved for $x_b$ as well.
$x_b=\frac{z}{12}-\frac{10}{12}x_a$. Now you set $z$ equal to $0$.
$x_b=-\frac{10}{12}x_a$. That means your objective function has the value $0$. This happens if $x_a=x_b=0$. Thus the line go through the origin with the slope $-\frac{10}{12}$. You need one more point to draw the line, for instance $P_2(3,-2.5)$. Now you have the objective function as a line with the initial level 0. This is the red line.
Then you push the red line right upwards parallel until the line touches the feasible region (green) the $ \texttt{first time}$. The blue lines illustrates the process of moving the line upward.
