what kind of function for graphing is it when you have an X with a tiny 3 near the top of it?
80482
This would most definitely be a polynomial function, since it starts with x cubed. Anything that has x cubed or higher in it, like x to the fifth power, would be a polynomial function. An equation with x squared would be a quadratic function. An equation with just x would be a linear equation.
775797
http://www.coolmath.com/graphit/
572199
I guess my favorite is sin(x). The graph of sin(x) is pure and simple. Starting at the origin like many other famous functions, it curves up to a maximum, curves down to a negative minimum, and returns to the x-axis. This graph is reflected everywhere in nature, from wheels, pendulums, and other physics-related concepts. Also, the area enclosed by one half of a sin(x) curve and the x-axis is 2. How cool is that! Sin(x) is better than all the other trignometric functions because it is the easiest to use.
108513
Asymptotes are vertical lines associated with a graph but not part of a graph.  They are divisions in a graph that I call "territory markers" or "fences".  They are essentially the x values that make any function undefined.  These x values, when graphed, are vertical lines that a graph will  not cross but rather be forced to "walk along".\n\nExample:  f(x)=1/((x)*(x-2))\n\nThe function is obviously undefined at 2 specific x values (set the denominator equal to zero and solve for those x vales to find where the function is undefined), x=0 and x=2.  The graph of the line x=0 and/or x=2 is the graph of a vertical line, or asymptote, that this function cannot cross directly since there is no corresponding y value at that point (the function is undefined at these points, remember?).  The function is forced to skirt the fence line either up to infinity or down to negative infinity.  Graph it and you'll see.
882816
You can't do that on a calculator. What you can do is use the calculator to get the graph drawn then use your algebra to figure out where to shade. \nLooking at the graph, you know that it's a parabola with the vertex (4, -3). test the point (0,0) outside of the parabola.....\nsince you get\n0 is greater than or equal to 0 - 0 + 13\nsince that is a false mathematical statement, you know that you shade inside of the parabola....\nlets test a point inside/on the parabola and see, like (4, -3)\n- 3 is greater than or equal to 16 - 32 + 13\n-3 is greater than or equal to - 3\nthat is true so you now know to shade inside your parabola\nthat's how you do problems like that w/a calculator\nsince your calculator can do all of the operations, it's really easy
140577
Buy a TI-89 graphing calculator, that'll do the trick.
372978
cool graph it little buddy
869140
Just graph it like an equality, and then your answer is everything either above (>) or below (<) the line.  For example, in the last inequality, the answer is everything above the line y = -2, not including the line itself.
158569
Here is a tutorial...you can change any of the parameters that you mentioned and look at the graph as it is made and changed...really cool if you ask me.\nhttp://www.analyzemath.com/logfunction/logfunction.html
729410
You can  find two different values for x that are easy to find 1/7 of, such as 7 and 14; see what y would be at those points [ -4, -3]; then plot the points in either order and connect the dots.  It is smart to also plot a third point to reduce the chance of error, because if one of the three is wrong, they won't line up.
922643
It depends on the graph. If the values of each notch differ by x, then the interval is x.\n\nJust look at two numbers on adjacent notches, and subtract.
41570
A parabola with its base at 0,1.  Go to http://www.coolmath.com/graphit/ Enter x^2+1 and see it for yourself!  :)
605811
As Lauren said, it's 1/x.\n\nMemorize this. It will be very useful.
402424
f+12<18  Subtract 12 from each side.\nf < 6\nThus, the solution is all numbers less than 6.\n(-infinity,6) is called interval notation.\nSet builder notation:\n{f|f<6}\nSet builder notation is always enclosed in braces {}, and always starts out with f|\nThe vertical bar is read "such that"\nThus, {f|f<6} is read "f such that f is less than 6."\nAlso, the graph is a number line with a parenthesis, ), at 6 and then shaded to the left of 6.
506906
In here if you just type x^2 everyone understands that to be x squared.\n\nIn MS Word, you can check superscript under edit font
794793
x = 3. Try entering it on your calculator.
830542
y = sqrt[x/(x-5)]\n\nStart by getting a few points:\nf(0) = 0  (it passes through the origin)\nf(5) is undetermined (divide by zero)\n\nSee what happens when 0 < x < 5.  For x = 3, x/(x-5) is negative, therefore the function is undefined over that range.\n\nFor x > 5, x/(x-5) is positive, so the function exists.  When x gets very large, x >> 5 (e.g., x = 1 million), x/(x-5) approaches one, so f(x) approaches sqrt(1) = 1 from above.  Therefore, y = 1 is an asymptote, but f(x) is always larger than one for very large x.\n\nNow look at x < 0.  x/(x-5) is always positive, and it's always less than one.  A good point to use is:\n\nf(-4) = sqrt[-4/(-4-5)] = sqrt(4/9) = 2/3\n\nand another good point is\n\nf(9) = sqrt[9/(9-5)] = sqrt(9/4) = 3/2\n\nWe know enough to plot the graph:\n\n1.  Draw the axes; draw a dotted-line asymptote at y=1; and draw another dotted-line asymptote at x=5.\n\n2.  Let's do negative x first.  Start at the origin, draw the curve through (-4, 2/3), and continue the line out to the left, using y=1 as an asymptote.\n\n3.  The function is undefined for 0 < x <= 5.\n\n4.  For x>5, start the curve way up on the x=5 asymptote, bring it down through (9, 3/2), and continue it out to the right, using y=1 as an asymptote.