Question 228691
If you want the maximum value of this function, there are a couple of ways to find it. And, when you get to CALCULUS, there will be an even EASIER way to do it!  For now, I'll show you METHOD I (Graphing Calculator) and METHOD II (Vertex at x=-b/(2a) method!) 


METHOD I:  Graph the function  y=-60x^2+ 240x with a graphing calculator (or the old-fashioned way by just plotting points, realizing that a graph in this form with the x^2 with a negative coefficient is a parabola opening downward!)


Of course, if you graph this with a standard window, it won't even look like a parabola--it looks like a line.  So you need a "larger" window on the calculator.  This is a trial and error process to decide what x and y values will give you the best looking graph.  You might want to start by finding the x-intercepts.  Let y=0, and solve -60x^2 + 240x = 0.  Factor the common factor of -60x, which leaves -60x(x-4)=0.  The x intercepts are at x=0 and x=4, so I recommend that you make your calculator x window go from, let's say, x=-4 to x=6.  Then notice that the graph goes way off the top of the graph.  So use larger values of y, like y=-100 to y=100.  The graph still goes way off the top of the graph, so it turns out that y=-100 to y=300 looks pretty good. Here is what your graph should look like:
{{{graph(300,300, -4,6,-100,300,-60x^2+240x)}}}


Now that you have the graph, you can probably see that the maximum point of the graph is at x=2.  If x=2, plug it back into the formula and get y=-60*2^2+240*2, which is y=240 for the maximum value.  You can also get this directly from the calculator using the TRACE button (if you have a TI83 or TI84!), or the "Maximum/Minimum" function if you know how to use that.  


By the way, if ANYONE needs help with a TI calculator, send me an Email at rapaljer@scc-fl.edu, and I'll send you (free, of course!) a link to the page on my website explaining TI graphing calculators -- I made it a LOT easier than that HUGE book that came with your calculator!


METHOD II:  Find the vertex using the {{{x=-b/(2a)}}} method.

It turns out that the vertex of a parabola {{{y=ax^2 +bx +c}}} is at the point where {{{x=-b/(2a)}}}.  


In this case {{{y=-60x^2 + 240x}}} so a=-60 and b=240, so 
{{{x=-b/(2a)}}}
{{{x=-240/(2*-60)}}}
{{{x=-240/-120=2}}}


If x=2, then {{{y=-60x^2+240x}}}
{{{y=-60*2^2+240*2=240}}}


Which method do you like??  Anyone can send me an Email at rapaljer@scc-fl.edu.


Dr. Robert J. Rapalje
Seminole State College of Florida
Altamonte Springs Campus