You can always graph by plotting some points! If x= 0, f(0)= 2. f(1)= -3+ 5+ 2= 4, f(-1)= -3- 5+ 2= -6 etc. Knowing that the graph is a "downward opening" (because the leading coefficient is negative) parabola should help drawing the graph.

A little more "sophisticated" method is to "complete the square". Write the function as f(x)= -3(x^2- (5/3)x)+ 2= -3(x^2- (5/3)x+ (25/36)- (25/36))+ 2 (A "perfect square" is of the form (x- a)^2= x^2- 2ax+ a^2. Here the coefficient of x is -(5/3) so 2a= 5/3, a= 5/6, and a^2= 25/36.)

Continuing, f(x)= -3(x- 5/6)^2+ (-3)(-25/36)+ 2= -3(x- 5/6)^2+ 49/12. Since a square is never negative, -3 times a square is never positive. f(x) is 49/12minussomething so this will be a maximum when that square is 0. That is, when x= 5/6, f(x)= -3(0)+ 49/12= 49/12. For anyotherx, f(x) is less than 49/12. That tells us that the "vertex" of the parabola is at (5/6, 49/12). Setting x= 0, as above, f(0)= 2 so the graph crosses the y-axis at (0, 2). Solving f(x)= -3(x- 5/6)^2+ 49/12= 0 gives (x- 5/6)^2= 49/36. Taking the square root of both sides, x- 5/6= +/- 7/6 so x= 5/6+ 7/6= 12/6= 2 and x= 5/6- 7/6= -2/5= -1/3. The graph crosses the x-axis at (2, 0) and (-1/3, 0).

Knowing that (1) the graph is a downward opening parabola, (2) the vertex is at (5/6, 49/12), (3) the graph crosses the y-axis at (0, 2), and (4) the graph crosses the x-axis at (2, 0) and (-1/3, 0) should let you draw a pretty good sketch of the graph.