Since #f(x)=0# when either #-x#, #2x+5#, or #3x-10# are equal to zero, #f(x)# has zeros at #x=0, -5/2,# and #10/3#.
So now we have 3 points to work with. To find out what happens at each point, we can just look at the different parts of the equation.
To find out what happens around #x=0#, just look at the #-x# part of the function. This is linear, so the function passes the point #(0, 0)# in a straight line.
To find out what happens around #x=-5/2#, just look at the #(2x+5)^2# part of the function. This is quadratic, and it's a perfect square, so there's a double root at #x=-5/2#. This means that the function just touches the x-axis at this point.
To find out what happens around #x=10/3#, just look at the #3x-10# part of the function. This is linear, so the function just passes through the point #(10/3, 0)# in a straight line.
We can also find out what happens at the two extreme ends of the function (the end behavior of the function). If #x# is really big (approaching #\infty#), then #f(x)# would be really small (approaching #-\infty#), since every part of the function would return a positive number, and there's a negative sign in front of everything. If #x# is really small (approaching #-\infty#), then #f(x)# would also be really small (approaching #-\infty#).
With all of this information, we can sketch the graph of the function:
The function starts out on the left side being really negative, and then goes towards the x-axis, but never actually crossing the x-axis at #x=-5/2#. It touches the x-axis, and goes back up (like the function #y=x^2#). At #x=0#, it passes through the point #(0 ,0 )# in a straight line (like the function #y=-x#). It then goes back towards the x-axis at #x=10/3#, passing through the point #(0, 10/3)# in a straight line (like the equation y=x).
You can check with the below graph:
graph{-x(2x+5)^2(3x-10) [-5, 5, -700, 700]}
OR, if you want a more simple way to graph the function, just plug it into your graphing calculator or WolframAlpha or any graphing utility.
