Using TI-83 to get the Parabola?
816484
You need to do two things:\n1. set your window\n2. program the equations.\n\n1. At the top there's a 'Window' button.  You can set window parameters there.  However, I'd recommend using the 'Zoon' button and chosing option 6: ZStandard.  This will give you a good window.   You can change it later.\n\n2. The top left button says 'Y='.  Press this and program equation 1 then 'enter'.  This put you down one line so that you can program equation 2.\n\nTo run press the upper right hand button 'Graph'.\n\nAlso, go to the Texas Instruments website for help and to download an users manual.
738843
I'd recommend graphing the function on a calculator (I use a TI-83) and use the 'trace' to find zeros.
65888
a parabola facing up!\n\nwith center (0,1)
699042
Get your formula into this format:\n\ny = (x+a)²\n\nWhen it's in this format, it's easy to get the axis of symmetry. The axis of symmetry is the line x = -a.  (This is also the value (x=-a) that will give you the vertex of your parabola.  The vertex and line of symmetry are related, because the vertex always goes through the line of symmetry.)\n\nSo since you have y = (x+2) ² your axis of symmetry is the line x = -2.\n\nHope that helped!\n\n~ ♥ ~
769871
1.  The function decreases on the interval [-infinity,4].  (Your function is a parabola opening upwards, with one bottom vertex at the point (4,0).)\n\n2.  This function is also a parabola opening upwards, with its vertex at (2,-4).  On the interval [-infinity,4], the local min. is at (2,-4) and there is no max.\n\n3.  You have local max at x = -13/3, and local min at x = 1\n\n4.   f(x) is increasing on [-infintity,1], decreasing on [1,5], and increasing again at [5,infinity]\n\nHope that helped.  Email me if something's not clear!\n\n.
892318
Take the right triangle to be arranged in the first quadrant.\n\nThe hypotenuse forms the line y = (-4/3)x + 4\n\nTake the dimensions of the rectangle to be x and y, subject to the above constraint.\n\nWe want to maximize A = xy.\n\nWe've solved for y above. This gives:\n\nA(x) = x[(-4/3)x + 4]\nA(x) = -(4/3)x^2 + 4x\n\nIf you're not taking calculus, graph this function and find the x that maximizes the function. It should be at the vertex of a parabola that points down - a clear indication that it is the maximum.\n\nIf you're taking calculus, find A'(x) and set it equal to 0.\n\nA'(x) = (-8/3)x + 4 = 0\n\nx = 4*(3/8) = 1.5\ny = (-4/3)* 1.5 + 4 = 2\n\nThe rectangle measures 1.5 x 2 cm, and has an area of 3 cm^2.
880047
I can find this on a TI-84+, is that how you want to get it or do you need an algebraic solution?
901756
You actually don't even need to know the intercepts.  Given that the parabola is symmetric about x = 4, you can just travel from (12, 4), through the line x = 4, and as far to the other side.  You need to travel 8 to the left from (12, 4) to get to x = 4, and if you travel 8 more left, you get to (-4, 4).
1147966
Yes!!!!!!!!! Email me, and I'll try to shed more light on this. Parabolas are where my genius comes out :)
133564
The coordinates of the vertex of a parabola \ny = ax^2 + bx + c is\n[-b/2a , (4ac - b^2)/4a]\nJust substitute and you will get\nV(2 , -7)
821185
I would complete the square first of all:\n\ny = x² - 4x = x² - 4x + 4 - 4 = (x-2)² - 4\n\nSo now your parabola's equation is in the format y = (x-a)² + b\n\nWhen it's in this format, you can easily tell the axis of symmetry, it's x=a.  So in our case, we see that our axis of symmetry is at x=2.\n\nTo get a table of values, just plug in various values of x to solve for y, as follows:\n\nx | y\n--------\n-1 | 5\n0 | 0\n1 | -3\n2 | -4\n3 | -3\n4 | 0\n5 | 5\n\netc.\n\nHope that helped!\n\n~ ♥ ~
232033
you would shade inside the parabola because it is >. you use a dotted line because it is > and not >=.
29226
Always Google! E.g., http://www.howardcc.edu/math/calculator/ti83frame.htm
1185644
u got it wrong\nx=at^2 and y=2at\n\nparabola y^2=4ax\n=> y^2 / 4a = x\n=> y^2 / 4a^2 = x/a\nlet this b equal to t^2\n\nthus \ny^2 / 4 a^2 = x/ a = t^2\nthus y= 2at and x=at^2
753539
Since f(x) is a parabola that opens upward, you need the minimum value, which is the y-coordinate of the vertex.  The x-coordinate is -(-12)/(2•3) = 2.  So the minimum y is at 3(2)² - 12(2) + 13 = 1, and your range is y >= 1.\n\nFor the second part, just solve 3x² - 12x + 13 = 28.\n\n3x² - 12x - 15 = 0\nx² - 4x - 5 = 0\n(x-5)(x+1) = 0\nx = 5 or x = -1.
844444
Remember that the vertex of the parabola for y = x² is at (0,0).  You can confirm this by putting x = 0 in and seeing that you get y = 0.\n\nIf you instead replaced x with (x - 3):\ny = (x - 3)²\n\nNotice how it would take a value of x = 3, to get y = 0.  In other words, the whole graph, including the vertex would move 3 units to the right.  The new vertex would be at (0, 3).\n\nSimilarly, if you replaced y with (y - 2):\ny - 2 = x²\n\nThis is the same as:\ny = x² + 2\n\nSo every value would move up 2.  The graph would move two units up and the vertex would similarly move to (0, 2).\n\nCase 1:\nParabola moves 3 units to the right -- vertex now at (3,0)\n\nCase 2:\nParabola moves 2 units up -- vertex now at (0, 2)\n\nExtra credit question for you:\nWhat would happen if you did both?\n(y - 2) = (x - 3)²
616029
For TI-83 or 84:\n\nHit "LOG" button to the left of the "7" button. \nType 1000. \nHit the ")" button.\nPress "Enter"\nThe number 3 should be displayed.\nThe "LOG" button on a calculator means log to the base of ten.\nThe "LN" button on a calculator means log to the base of the natural number, e.