For the equation y=ax squared+bx+c....?
916256
first express the equation in the form\ny=a(x-h)^2+k\nif the a is positive itwill open upwards\nif a is negativeit will open downwards\nfor finding the y intercept put x=0\nin this form y=ax^2+bx+c y intercept will bw (0,c)\nthe vertex will be given by(h,k)
929562
∫ax(2 -x)dx\n= ∫(2ax - ax²)dx\n= 2a * ½x² - a * ⅓x³ + c\n= ax² - ⅓ax³ + c
600230
Do you mean (x-h)^2? If so, yes, this complies with the general equation of a parabola y = ax^2+c. \nThe value of y is proportional to the square on x, that has to be a parabola.
316569
y = mx + c\n(0) = (2)(-1) + c\nc = 2\n\ntherefore y = 2x + 2
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)
1051508
Your equation would look like x^2-ax+b.  Which means when you solve the equation you would two answers one of which will contain a plus sign other will contain a minus sign since the middle term is negative.  Thus the answer is C.
21539
If you have a system of equations with two variables of the form ax + by = c, dx + ey = f, where a,b,c,d,e,f are the coefficients, and x and y are variables, you can set up the equation as \n\n[a b][x] = [c]\n[d e][y] = [f]\n\n(ignore the fact there are two equals signs there, they're only there for formatting reasons)\n\nwhere the square brackets are a crude attempt on by behalf to 'draw' matrices and column vectors.\n\nTo solve the system you need to find the inverse of the matrix of the right hand side (assuming it's invertible to begin with), and "right multiply" (that is multiply from the right of each side of the equation) both sides of the equation.\n\nThat should give you:\n\n[1 0][x] = [ inverse of matrix ][c]\n[0 1][y] = [ ........................ ][f] \n\nThen it's just a matter of performing the calculation.\n\n(sorry for the crappy formatting of the post)
920068
The standard form of a quadratic:\n\nax^2 + bx +c\n\nIn your case a =1, and c=1. Since there is no x term, b=0
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
965380
(x - 3)² + (y + 2)² = 4²\n(C): x² + y² - 6x + 4y - 3 = 0
739779
ax²+bx+c=0\nDivide by a:\nx² + bx/a + c/a = 0\nSubtract c/a:\nx² + bx/a = -c/a\nAdd (b/(2a))²:\nx² + bx/a + b²/(4a²) = -c/a + b²/(4a²)\nFactor and simplify:\n(x+b/(2a))² = (b²-4ac)/(4a²)\nTake the square root:\nx+b/(2a) = ±√(b²-4ac)/(2a)\nSubtract b/(2a):\nx = (-b±√(b²-4ac))/(2a)
30454
say the line is y= ax + b\nsubstitute the (x,y)= (3,-11), (-1,9)\n\n-11= 3a+b......(1)\n9= -a+b.......(2)\nsubtract (2) from (1)\n\n-11-9= (3a+b)-(-a+b)\n-20= 4a\na=-5\nsubstitute this back into (2)\n9=5+b\nb=4\n\nie the original equation is y=-5x+4\n\nThe answer is D. but u left "x" out
429058
x==(-b+/-sqrt(b^2-4ac))/(2a).
482048
-b (+/-) sqrt(b^2 - 4ac)\n--------------------------------- = x\n2a\n\nUsing the equation "ax^2 + bx + c = 0" to find the values of a, b, and c.
954269
it is natural to write the equation of st.line as y=ax+b (now common form is y=mx+c; this has its own story!)\n\nso if x=0, y=b
617358
x = (-b +- sqrt(b^2-4ac))/(2a)\n\nThis equation solves the quadratic equation, where\n\nax^2 + bx + c = 0.
231877
ax^2+bx+c=0\nx^2+3x-9=0\n\nformula:\nfirst root:\n(-b+sqrt(b^2-4ac))/2a==> (-3+sqrt(3^2+4*9))/2=1,85\nsecond root:\n(-b-sqrt(b^2-4ac))/2a==> (-3-sqrt(3^2+4*9))/2=4,85