how to solve graph of quadratic funtion???
91768
Quadratic Functions have the form y = ax^2 + bx + c\n\nThe vertex will always have x = -b/(2a). You can calculate this number and plug it into your equation to solve for y. If a is positive, the graph will go up in both directions from the vertex. if a is negative it will go down. You can pick some x value different from the vertex to help you plot other points to allow a symmetric parabola graph.\n\nGood Luck!
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)
825543
ok , \nyou need to plot a graph of the equation 2x^2-4x-3 .\nto b more clear: 2*x^2 - 4*x - 3\n\nFor that u need to take the horizontal axis as x = 0,1,2,3,...\nand vertical axis (y axis ) as 2x^2-4x-3\n\nFor every value of x obtain a value for y,\neg.\nfor x = 0 ; 2x^2-4x-3 = 2 * 0^2 - 4*0 - 3 = -3\nfor x = 1 ; 2x^2-4x-3 = 2 * 1^2 - 4*1 - 3 = 2 - 4 - 3 = -5\nfor x = 2 ; 2x^2-4x-3 = 2 * 2^2 - 4*2 - 3 = 8 - 8 - 3 = -3\n.\n.\n.\n.\nand so on for as many points you need to plot\n\nnow with values of y as -3,-5,-3 corresponding to x = 0,1,2 plot appropriate points and join them to obtain the graph.\n\nto plot a point , take the first point (0,-3) thats for x = 0 , y = -3,\nand so on (1,-2) & (2,-3).\n\nI hope this helps :)
361536
Let's call f(x) = x^2 - 4\n\nThe graph of f(x) = x^2 - 4 looks like a smile ( U )  with the graph passing the y - axis at -4.\n\nThe roots are at x = 2 and x = -2.  This can be found by solving for x^2 - 4 = 0\nx^2 - 4 = 0\n(x + 2)(x - 2) = 0\nso x + 2 = 0, or x - 2 = 0\nso x = -2 or x = 2\n\nsince you're int'ing from x = -1 to x = 3, you have to break up the int into\n[x = -1 to x = 2, where f(x) is negative and take the absolute value] \nand\n[x = 2 to x = 3, where f(x) is positive]\n\nso int of f(x)\n= x^3/3 - 4x + c\n\nso int of f(x) from x=-1 to x=3\n= {+ve value of int of f(x) from x = -1 to 2}\n                +\n  int of f(x) from x = 2 to 3\n\nSeparately\nint of f(x), x = -1 to 2\n= | (2^3)/3  - 4(2) - [((-1)^3)/3  - 4(-1)] |\n= | (8/3) - 8 - [(-1/3) + 4] |\n= | (8/3) - 8 + (1/3) - 4 |\n= | (9/3) - 12 |\n= | 3 - 12 |\n= | -9 |\n= 9\n\nint f(x), x = 2 to 3\n= (3^3)/3  - 4(3) - [(2^3)/3  - 4(2)]\n= 27/3 - 12 - [(8/3) - 8]\n= 9 - 12 - (8/3) + 8\n= 5 - (8/3)\n= 7/3 (or 2 1/3)\n\nAdd them together and you get:\n9 + 2 1/3\n= 11 1/3 which is appromixately 11.333_
846578
1.  Since f(x) = x² + bx + c, and you know f(4) = 9 (since (4,9) is a point on the parabola, specifically the vertex), you can solve for f(4) = \n\n16 + 4b + c = 9, so 4b + c = -7\n\nSince (4,9) is the vertex, you know the tangent at this point is zero, i.e., f '(4) = 0.  (I hope this is a calculus HW problem so that you know what I am talking about!)\n\nSo get f '(x) = 2x + b\n\nSince f '(4) = 0, we have 2(4) + b = 0, so b = -8\n\nSince b = -8, and 4b + c = -7, we can now solve for c:\n\n4(-8) + c = -7, so c = 25\n\n----------\n\n2.  Not sure I can help here.  I took the first and second derivatives of h(x) to determine that the function has a local min at x=7/3 and a local max at x=-1, but h(7/3) ≠ -5 and h(-1) ≠ 2, so not sure where to go from here, sorry!\n\n----------\n\n3.  A parabola has a 2nd-degree polynomial equation, or y = ax² + bx + c.  We know two values, namely (4,4) and (2,8) so we can set up two equations like thus:\n\n4 = a(4²) + b(4) + c\n8 = a(2²) + b(2) + c\n\nwhich we can rewrite as:\n\n4 = 16a + 4b + c\n8 = 4a + 2b +c\n\nRight now we have 3 unknowns and only 2 equations, but we can get a third equation by taking the first derivative, or f '(x) = 2ax + b.  Since the vertex is at (4,4) we know that f '(4) = 0, and we can get a third equation:\n\n0 = 2a(4) + b = 8a + b\n\nSo now we have 3 equations in 3 unknowns and we can solve for a, b, and c:\n\n4 = 16a + 4b + c\n8 = 4a + 2b +c\n0 = 8a + b\n\nSubtract equation 1 from equation 2, then divide by 2 to get:  \n2 = -6a - b\n\nNow add equation 3 to this to get 2 = 2a, or a=1.\n\nNow let's back-substitute to get b and c:\n\n2 = -6(1) - b, so b = -8\n\n8 = 4a + 2b +c\n8 = 4(1) + 2(-8) + c, so c = 20\n\nSo the equation of our parabola is:\nf(x) = x² -8x + 20\n\n(Double check to make sure f(4)=4 and f(2)=8).
778614
y =mx + b\n\nFirst, the word linear will tell you this is the answer, as linear graph functions follow this format. From "best expressed", i got the hint that 4) is the answer cos its is the most direct relation. i.e. you plot in x and you get y, like a direct mirror.\n\n1) A quadratic equation will give you 2 solutions to a number you plot in, and from this itself, we can eliminate it. \n2) Complicated. The relationship isnt clear also, cos the entire thing seems to be pure gibberish\n3) This formula can only be applied in a right angled triangle. So it only shows ONE kind of relationship two variables can have.\n\nHope this helps=)
1068232
Use quotient rule to get the derivative, which, simplified, is:\n\n-30*(2x+3)/((3x^2+9x+7)^2)\n\nSet it equal to zero and solve.  Since the denominator cannot be zero, set the numerator equal to zero\n\n2x+3=0\nx=(-3/2)\n\nI checked it, and it indeed is the solution. \n\nBy the way, you don't have to worry about an undefined points of this function making problems in the graph or in the max/min values in the derivative.  The denominator has imaginary roots.
921939
Good Evening\n\nI think we've been avoiding this question because we can't graph here... however to get you started, create an x|y chart for each line, plug in a couple of values, and connect the dots  (for a line you only need two points, but some books recommend finding three, so you can verify you didn't make a mistake somewhere)\n\nx+y = 8\nx|y\n----\n4|4\n3|5\n2|6\n\nx - y = -4\nx|y\n0|4\n1|5\n2|6\n\nGraph the two lines, and they should intersect at (2,6) since it appears in both of the charts...\n\n~i~
439883
Use the following four steps to solve a quadratic equation, in this order:\n\nStep 1: Guess the answer\n\nIf x^2=0, can you guess the answer?\nIf x^2=1, you can guess, but remember that there may be two answers. (1 and -1)\n\nYou can tell if you are guessing right by plugging back into the original equation. If it worked, you're done.\n\nIf you can't guess in 30 seconds, you are done with Step 1. Go to Step 2.\n\n\nStep 2: Try factoring\n\nx^2 + 4x + 4 = (x+2)(x+2)\n\nRemember that the constant term of your factors has to multiply to the constant term of the quadratic. \n\nFor example, if you have x squared plus something plus 6, remember that your constant terms must be multiples of 6. \n\nTry:\n\n(x + 1)(x + 6)\n(x - 1)(x - 6)\n(x + 2)(x + 3)\n(x - 2)(x - 3)\n\nYou just ran out of factors of 6. If this didn't work, go to step 3.\n\n\nStep 3 -- Complete the Square\n\nOne important thing to remember about completing the square is that it does exactly the same thing as the quadratic equation. Pick either Step 3 or Step 4 as your last step. Guessing and factoring don't always work because you can't always guess and you can't always factor. But completing the square and using the quadratic formula will always find the answer if there is one.\n\nSuppose you have \n\nx^2 - 6x - 16 = 0\n\nHere's how to complete the square:\n\n#1. Move the constant (whatever it is) to the right side\n\nx^2 - 6x = 16\n\n#2. Find half of the coeficient of the x term and add the square of that number to both sides of the equation:\n\nThe x term here is -6x\nThe coefficient is -6\nHalf of -6 is -3 (Remember this number -3)\n-3 sqared is 9. Add 9 to both sides of the equation:\n\nx^2 - 6x + 9 = 16 + 9\n\n#3. If you did #2 correctly, then the left side is a perfect square and the constant term is -3 (remember -3?)\n\n(x - 3)^2 = 25\n\n(If you're not sure, multiply it out and see if you get what you had before.)\n\n#4. Take the square root of both sides. Remember, it could be negative or positive.\n\nx - 3 = 5\nx - 3 = -5\n\n#5. Solve your two equations\n\nx = 8\nx = -2\n\n#6. Put your answers back into the original equation and see if you got the right answer. \n\n\nStep 4. \n\nIf you did Step 3, you already have the answer, and you're already done. If you didn't do Step 3, use the quadratic formula to get the answer. You can find the quadratic formula here:\n\nhttp://en.wikipedia.org/wiki/Quadratic_formula\n\nMemorize the quadratic formula and never forget it.  Write it on your forhead until you have it memorized.
597648
y= -4x^2+8x-1\ny = -(4x^2-8x+1)\ny = -[(2x-2)^2 - 4 + 1]\ny = -[(2x-2)^2 - 3]\ny = 3 - (2x-2)^2\n\nSo the vertex is (1,3) and it is a maximum point.
736099
plot the coordinates then do the vertical line test.  if the vertical line intersects more than two points on the graph then it is not a function.  each set of coordinates you have listed are functions (i free hand it with no graphing paper or calc) so forgive me if im wrong.  also refer to www.coolmath.com. it is a good site for breaking down math
1177675
The best way is (I'm looking at my TI-85) to use the POLY function. (On the TI-85, this is the second function of PRGM.) When you open this function, it'll ask you the order of your polynomial function (I think it can handle up to 20). You enter it and then type in all the coefficients. It'll return the roots of the equation for you, real and imaginary.\n\nIf your calculator doesn't have the POLY function (or you can't find/operate it), another way would be to plot the graph and use the x-intercept tool to find the intercepts.\n\nGood luck! Message me if you have any other math/calculator troubles!
333191
Consider the function f as mapping the points x of a set X to the points y of a set Y. Draw a horizontal line (the x axis) so that each point on the line represents an element of the set X. Draw a perpendicular line (the y axis) to intersect the x axis at the null point or zero. Each point on the y axis represents an element of the set Y. \n\nFor each x in X calculate the value of y= f(x) in Y. This gives a pair of co-ordinates (x,y). (Depending on the function, you may have more than on value of y for each x eg the function f(x)=square root of x will give + and - values.)\n\nFor any x on the x axis draw a line parallel to the y axis to intersect the x axis at x. Draw another line to intersect the y axis at the point y=f(x) and parallel to the x axis. Where these lines intersect is a point on your graph. The complete graph consists of all such points.\n\nWhere the set X is infinite (eg the set of real numbers) you cannot calculate the value of f(x) individually for every x but by calculating the values for a series of points at suitable intervals you will see a pattern emerging to the graph and you will be able to fill in the gaps by joining the (x,y) points, hopefully with a smooth and continuous line.
560642
My dear, just go to www.quickmath.com and on the left hand side click on 'Equations' under 'Graphs' and thewn type in your equation u will get the graph. I have been using it for years....It's a great site!! I hope this helps :)
890433
go here- http://people.hofstra.edu/staff/steven_r_costenoble/Graf/Graf.html\n\nand enter     2x^2   in the big blank field and a graph will pop up!
608711
i think it's a parabola.. im not sure. u should try plotting it on a graph.. substitute  the x and y values and see what form comes out..
808711
Welcome!!\n\nHere is the solution:\n\nlog(x+1) = log3 - log(2x-1)\nlog(x+1) + log(2x-1) = log3\nlog[(x+1)(2x-1)] = log 3\n\ntherefore,\n\n(x+1)(2x-1) = 3\n2x^2 - x + 2x -1 = 3\n2x^2 + x - 4 = 0\n\nyou need to use the quadratic formula to solve this:\n\nx = {-1 ± sqrt[1 - 4(2)(-4)] } / 2*2\nx =[ -1 ± sqrt(33)] / 4