Question 413891
maximum/minimum will be turning point.  Differentiate the function {{{y=((1/2)x^2)-x-4}}}  which is what I think you mean gives you a gradient of x-1, so will have a minimum (minimum as is positive {{{x^2}}} so a bowl shape with a minimum and no maximum) at x=1, at which point y = -4.5.

zeros;
x=0, y=-4 clearly
y=0, {{{x^2 -2x -8 = 0}}}   - I multiplied by 2 to make easier
0=(x-4)(x+2), x= 4 or -2

Line of symmetry I am fairly sure it doesn't have as it is a function of x which is an odd function, and x^2 which is even.7

Drawing the graph should be fine now you know where it crosses the axes.