How do I solve a quadratic equation by graphing?
156411
If you are solving f(x) = ax^2 + bx + c = 0 by graphing, then there are three possibilities:\n\n1) graph of f(x) does not touch x-axis. Then there are no real solutions. Your real graph (x and y axis) does not tell you the solutions. You will have to do this algebraically (plug a, b, and c to quadratic formula). you will notice that you have to take the square root of a negative number. that is why these roots are not real numbers. they are both complex numbers.\n\n2) graph of f(x) touches x-axis at one point. Then there is one solution (with multiplicity of 2) at this x value. also, the quadratic polynomial factors into a perfect square.\n\n3) graph of f(x) touches x-axis at two distinct points. you can find the two solutions by their values on the x-axis.
756992
Don't you just love the quadratic equation?\n\nThe first thing you got to do is get the given equation into the correct form, and in order to do that you have to combine all like terms. Since there are no like terms yet you have to do some math first.\n\n-2.0^2 = 4.0because neg * neg = pos\n\nNow you have 2 like terms 4.0 and 0.5 combine to get 4.5. Your eqution should look like this:\n\nf(t)=10t + 4.5 since you need at^2 + bt^2 + c just insert t^2. Now your equation will look like this:\n\nf(t)=t^2 + 10t + 4.5\n\nAnd just in case you need the quadratic equation:\n\nt= -b + or - sqr(b^2 - 4ac) / 2a\n\noh yeah and a = 1
812734
The easiest way to graph this equation is to plug in 0 for x and get one point, then plug in 0 for y and get another, then draw the line that connects the two points.\n4(0) - 3y = 24\n-3y = 24\ny = -8\n(0, -8)\n\n4x -3(0) = 24\n4x = 24\nx = 6\n(6, 0)
145305
Use cubic equation \ny=ax3+bx2+cx+d\n\nSince you have four points data you can get four linear equations with four variables which is very easy to solve, get variables (a,b,c,d) and plug the values to cross check and see u will get answer. Use guass jordan or elimination method.
778230
y = ax^2 + bx + c, a not equal to 0\n\nvertex = [(-b/2a), f(-b/2a)]\n\n-b/2a = -4/2 = -2\n\n-2^2 + (4 x -2) + 5 = 4 - 2 + 5 = 7\n\nthe vertex of the graph of the equation is -2, 7\n\nTherefore the axis of symmetry is x = -2
781653
Eliminate the y to get\n2-x^2=2x+4\nx^2+2x+2=0\nusing quadratic formula\n(-2+-sqrt(4-4*2))/2\nwe see there are no real roots: the two functions do not intersect\nx= -1+-i
121228
here is a simple program algo\n\nprogram quad\nbegin\nvar\n     declare variables;\nget variable values of a,b,c for equation ax2+bx+c=0\nd=)b*b)-4*a*c;\nif(d<0)\nbegin\n\n   root are imaginary;\nend\nelse\n\n    root1=b(sqrt(d))/2a;\n    root2=-b(sqrt(d))/2a;\n ( note : formula i am not sure make changes as per required)\nprint the value of root1 and root2;\nend
809000
There are many ways to approach this question the easiest being if you have a graphing calculator just plug in a few points ie. (23,9200) (24, 9216)<look below for explanation> into stats and then use the stats operation quadreg and plug that formula into your y values and look at the table for the biggest y or you can look at the graph and use the maxium feature.\n\nif you don't have a graph calc then you need to grab a few points ... I'll use (23,400) (24, 384) (25, 368) (26,352) and (27, 336)... now multiply the x and y for each and then put that in the y value for each x.... ie. (23,9200) (24, 9216) (25, 9200) (26, 9152) (27, 9072)\n\nmake make a small t-chart (with 2 extra columns the first d and then d') with your first two as your new x then y points.... with the first extra column(d) take the difference of each y(for the first entry you will not be able to have an answer) ie. 16; -16; -48; -80; now in the second extra column(d') take the difference of the d column and multiply it by -1 ... -32, -32, -32\n\nnow you ready to start making your quadratic... take d'(32) and take it to the -2 power<I may be wrong here and it might be the -1 power, but you can just test and come back if wrong> then place that number in front of x^2 (I will refer to this number as a)\n\nok now take this formula ax^2 +bx +K(this is the constant but in this problem you have none) now plug in a value you have for x from your second points and solve for b now this is your quadratic, you should test it on a point or two, but use this to create a huge t-chart of different points and look for the when the value gets smaller, and the point before it gets smaller is your point\n\nwe actually ran into the answer while doing the problem which is common in math books but the point (24, 9216) or when you have 24 trees is the maximum yield
725212
Yeah, make them graphable on your calc.\n\ny = 4-x\ny = 2x-2\n\nIf you want to solve, just set 4-x and 2x-2 equal to each other (since y must also be equal in both equations)\n\n4-x=2x-2\n3x=6\nx=2\n\n\n2+y=4\ny=2\n\nTheir intersection is (2,2)\n\nThe y-intercept of the first is 4, and the second is 2.\nThe slope for the first is -1, and the second is 2.\nThe x-intercept of the first is 4, and the second is 1.
741586
set y=0\nsolve quadratic eqn\nquadratic eqn\n\n-7 +- (49-16)^.5\n-------------------\n2\n= (-7 +- 33^.5)/2
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 :)
749496
By using the equation -b/2a you can get the x coordinate of the vertex and then plug that answer into the original equation to fond the y component.\n\ny=ax^2-bx+c, -b/2a\ny=2x^2-8x+15, -(-8)/(2*2)=2, plug 2 in for x and y=7. So vertex is at (2,7)
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!  :)
939068
Is this one problem?  If so, you can solve it like this:\n\n(a/b) / (c/d) = (a/b) * (d/c)\n\n~ ~ ~ ~ \n\n( (x² - 4) / x² ) / ( (x²+5x+6) / x ) = \n\n( (x² - 4) / x² ) * ( x / (x²+5x+6) ) = \n\n( (x² - 4) * x) ) / (x² * (x²+5x+6) ) = \n\n( (x² - 4)  / (x * (x²+5x+6) ) = \n\n{{ now notice you can factor your quadratic equations, to get the following }}\n\n(x+2)(x-2)  / (x * (x+2)(x+3) ) = \n\n(x-2) / (x(x+3))\n\nHope that helped!\n\n~ ♥ ~
831674
1. replace x with -3, calculate and see if you get 0\n2. divide by (x+3) and solve the remaining quadratic equation
202779
This is quadratic. A quadratic equation looks like:\nax^2+bx+c = 0\n\nSo a=4, b=-1, and c=3\n\nSo by using the quadratic equation: \n\nx= (-b+sqrt(b^2-4ac))/2a\nand\nx= (-b-sqrt(b^2-4ac))/2a\n\nTry to get the answer from that.... note that imaginary numbers may be involved!
893823
Use the quadratic equation \n\nt = -b + or - sqr(b^2 - 4ac)/2a\n\nyour equation is already in the form:\n\nat^2 + bt + c = 0\n\njust plug in the variables and solve for t twice