Find the x-intercepts. y= x^2 - 7x + 7?
930634
x-intercept is where the curve crosses the x-axis\n\nequation of x-axis is y=0\nseeking the zeroes of x^2-7x+7\nthe values of x that make x^2-7x+7 zero.\n\n\nx^2-7x+7 = 0\nx = (7+/- sqrt(7^2-4.1.7))/2\n= (7 +/- sqrt(21))/2\n\nso the x-intercepts are\n(7+sqrt(21))/2,0) and (7-sqrt(21))/2,0)
431677
If you're dealing with a straight line (and a few other types of line I think), the equation will be of the form\ny=mx+c\nThe c part will give you the y-intercept i.e. where the line crosses the y-axis.
744326
yes!  \n\n2x-3y=7\n\n-2x from each side of the =\n\n-3y=-2x+7\n\n-3y/-3 and -2x+7/-3\n\ny=2/3x+-2.3\n\nGo to -2.3 on the Y-intercept\nThe slope is 2/3 (up 2, over 3)
804813
(y+1)(y-7)\n=y^2-6y-7
172343
the easiest way to find a y-intercept is to just plug in 0 for x.\n\nSo, you get:\ny-3=2(0-5)\ny-3=2(-5)\ny-3=-10\ny=-7
831999
y = (x+3)(x-1)(x-2) = x^3 - 7x + 6
993104
take the equation x^2 - 7x + 12 and do the following  (x ) (x) now...see that there is a - and a + in the equation....you need to do this ( x -  ) ( x - ) 'cause you put the minus in the first factor in the second factor you take the - and the + and multiply them....So you get (x- ) (x -) the other 2 terms for each equation comes from the 7x and the 12.  What you do is to find 2 numbers that multiplied gives you 12 and added give you 7....The numbers are 3 and 4 ...Then put them into your factors (x-3)(x-4).\n\nFor the other 2 equations you do the same process\n\n(x^2 - 8x + 7) = (x-  ) (x-  ). In this one, you have to find 2 numbers that multiplied give you 7 and added give you 8. They are 1 and 7. Then (x^2 - 8x + 7) = (x- 1 ) (x- 7 ).\n\nw^2 - 8w + 15 = (w - ) (w- ). As you saw in one and two, three is the same....so the factors are 3 and 5. Then, w^2 - 8w + 15 = (w -3 ) (w-5 )\n\nFor the solve...\n1. x^2 - 6x = 0\n2. x^2 - 2x - 3 = 0\n\nTHe 1 is like this...you have to complete your quadratic equation then you do the following x^2 - 6x = 0 --> x^2 - 6x + 9 =0+ 9...The 9 in the right side was added in order to keep the two sides equal...IN the left side you got a perfect quadratic equation, hence it will be (x-3)^2= 9....Now you pass the 9 of right to the left ...then it will be [(x-3)^2]-9=0\n\n\nThe other equation x^2 - 2x - 3 = 0....This one is quite simple, it is the same as the part 1...The difference is that when you find the factors, they will be 2 numbers that multiplied give you 3 but substracted gives you 2.....The #s are 3 and 1....Here is the thing....when you break the equation into factors, you will have the following (x- ) (x+ )...Since you have a -2 that means that it is the -3 and the +1...Hence your factors will be the following (x-3)(x+1)
839343
x^2 + 4x + 4 = 7 (subtrat 7 from both sides)\n\nx^2 + 4x -3 = 0 (use the quadratic equation formula)\n\n-b + or - the square root of (b^2 - 4ac) / a^2\n\n-4 + or - the square root of 16 - (4) (1) (-3) / 2\n\n-4 + or - the square root of 28 / 2\n\n-4 + or - (the square root of 4) (the square root of 7) / 2\n\n-4 + or - 2 (the square root of 7) / 2 (simplify by dividing 2 into  the -4 and 2 on the top to get rid of the 2 on the bottom)\n\nthat gives you x = - 2 + or - the square root of 7.
845468
Y-intercepts :   .5\n\nX-intercept:   1\n\nAssymptote at x = 2
852759
3x+7y=3\n7y=-3x+3\ny=-3/7*x+3/7\nSlope=-3/7\nWe have the slope and point[-6,7]\nWe will find the intercept\n7=-3/7*-6+c\nc=7-18/7\n=[49-18]/7\n=31/7\nWe now have the slope and intercept\nThe equation is\ny=-3/7*x+31/7\n7y+3x-31=0
863170
y=x^3-3x+1\nat x=0 y=1\nso y intercept is (0,1)\nfor x intercept set y=0\nx^3-3x+1=0\nsolve\ny'=3x^2-3\nsetting 3x^2-3=0\n3(x^2-1)=0\n(x+1)(x-1)=0\nlocal maximum/minimum at x=-1,x=+1\nf''(x)=6x\n-6 at x=-1 so max at x=-1\n+6 at x=1 so min at  x=1\n\nthe function is increasing from - infinityto -1\ndecreasing from -1 to +1\nincreasing from 1 to infinity\ncurve concave downwards at x=-1 and concave upwards at x=1
932643
4x + 7 = -7\n4x = 14\nx = 14/4\nx = 7/2
918456
domain\nall real x\nx intercept\nputting y=0\n0=x+rt{x]\n-x=rtx\nx^2=x\nx(x-1)=0\nx=0 or 1\nx intercept=(0,0) and (1,0)\ny intercept\nsetting x=0\ny=0\ny intercept=(0,0)\nno symmetry exhibited
805022
let the numbers be x and y\nSo x + y = 7\nxy = 25\n\nNow 1/x + 1/y \n= x/xy + y/xy\n= (x + y)/xy\n= 7/25\n\nx + y = 7 So y = 7 - x\nxy = 25\n\nSo x(7 - x) = 25\n\n7x - x² = 25\n x² - 7x + 25 = 0\nx = [-b ± √(b² - 4ac)]/(2a)\n= [7 ± √(7² - 4.1.25)]/(2)\n= [7 ± √(- 51)]/(2)\n\nSo x = [7 - √(- 51)]/(2) and y = [7 + √(- 51)]/(2) \nThus 1/x = 1/[7 - √(- 51)]/(2)\n= 2/[7 - √(- 51)] * [7 + √(- 51)]/[7 + √(- 51)]\n= 2[7 + √(- 51)]/(49 - -51)\n= [7 + √(- 51)]/50\n\n1/y = [7 - √(- 51)]/50\nSo 1/x + 1/y = [7 + √(- 51)]/50 + [7 - √(- 51)]/50\n= 14/50\n= 7/25 (as above but with much more effort!!!!!!!!) (8^O)
954948
a) Algebraically the x-coordinate of the vertex is x= -b/(2a)\n\nHere that is x = -10/(2 X 1) = -10/2 = -5\n\nPlug that in for x to get the y-coordinate of the vertex.\n\nb) To get the x-intercepts, replace y (here, g(x)) with 0 and solve.\n\n0 = x^2 + 10x + 21 = (x+7)(x+3)\n\nx = -7 or -3 so there are two x-intercepts: -7 and -3.\n\nc) to find the y-intercepts, replace x with zero. \nSo y = 0 + 0 + 21 = 21  \n21 is the y-intercept.
660785
i don't fucking know.\n\nEDIT: waitaminnit i'm gonna give it a try!\n\nx^3-4x^2-7x+10 = 0\n\nfor x = -1\n-1 -1 +7+10 != 0\nso x != -1\n\nfor x = 1\n1-1-7+10 != 0\nso x != 1 either!! ...goshdarnit.\n\nx^3-4x^2-7x+10 = 0\nx(x^2-4x) - 7x = -10\nx^2-4x - 7 = -10/x (we divided both sides by x)\n..ok...i think i've lost the plot sorry.
820064
x^2+2xy-y^2=7\n\ndy/dx\n2x+2y+2x(dy/dx)-2(dy/dx)=0\n(2x-2)(dy/dx)=-2x-2y\n\ndy/dx= (-2x-2y)/(2x-2)\ndy/dx=-3\n\nd^2y/dx^2\n[(2x-2)(-2-2(dy/dx))-\n(-2x-2y)(2)]/(2x-2)^2\n\nd^2y/dx^2=5\n\ni think thats rite sry bout all the parantheses