find the x and y intercepts of y = -4x-5?
399118
y-int= (0,-5)\nx-int= (-1.2,0)
813674
To find the y intercept make x equal 0 then solve the equation. You can then work out the x intercept by making y equal 0. This will allow you to plot the line and work out the gradient/slope by using rise divided by run. Alternatively, rearrange the equation to make y the subject. I believe the slope would be -1/4. This would be y=-1/4x+8.
1142461
1. x-y=5; y= x - 5\nAdd y to each side x - y + y = 5 + y, becomes x = 5 + y. Subtract 5 from each side, x - 5 = 5 - 5 + y, becomes x - 5 = y\n\nSolve the rest on your own
444240
-y=2-x\n\ny=x-2\n\nslope is 1\ny intercept is (0,-2)
523941
x intercept y=4 (0,4)\n(plug 0 in for x and solve)\ny intercept x=20/3 (20/3,0)\n(plug 0 in for y and solve)
957836
4x+(-5y)+(-6x)+3y\n4x+(-6x)+(-5y)+3y\n=-2x-2y
1136858
y + 1 = 2y + 5\nSubtracting y on both sides\n1 = y + 5\nsubtracting 1 on both sides\n0 = y + 4\nTherefore y = -4
418521
The Y-INTERCEPT of the line y = b is b.\n\nFor a y-intercept, you plug in 0 for the other variable, primarily x, but in this case, b.\n\nTherefore, if you plug it in, you receive:\n\ny = 0\n0 = 0\n\nThe y-intercept is 0\n\nPerhaps you should check by graphing, but I assure you that I am correct
203791
The x-intercept is when the line crosses the x axis. When your equation's graph crosses the x axis, it means that y is zero at that point. right?\n\nSo to find the x-intercept you assume y=0 and solve for x.\n\nNow to find the y-intercept, it is when x=0. If you imagine it graphically, when it touches the Y-axis, X's value is zero.\n\nSo to find the y-intercept you assume x=0 and solve for y.\n\nExample: y= 3x-6\nx-interp: assume y=0\n      0= 3x -6      \n 3x = 6     \n x = 2\n\ny-interp: assume x=0 \n     y= 3*0 -6        \n          y= -6
293210
the graph of |x - 2| + |y - 5| = 5 consists of four lines which are graphs of the following functions:\nx-2+y-5=5 which is  line A: x+y=12\n-x+2-y+5=5 which is line B: x+y=2\n-x+2+y-5=5 which is line C: x-y=-8\nx-2-y+5=5 which is line D: x-y=2\nnext thing to do is graph these four lines. the easiest way is to find the 0s. \nline A: x+y=12\n        x=0 y=12 point A1(0,12)\n        y=12 x=0 point A2(12,0)\nline B:  x+y=2\n         x=0 y=2 point B1(0,2)\n         x=2 y=0  point B2(2,0)\nline C: x-y=-8\n         x=0 y=8 point C1(0,8)\n         x=-8 y=0 point C2(-8,0)\nline D: x-y=2\n        x=0 y=-2  point D1(0,-2)\n        x=2 y=0   point D2(2,0)\nthese four lines enclose a figure that is a rectangle. lets find the length of each side. after you plot the four lines you will notice that line x+y=12 intersects with line x-y=2 lets call the intersection point I1 to find I1 coordinates we rearrange y=12-x and y=x-2 and from y=y we get x-2=12-x or 2x=14 or x=7 if x=7 y=12-7=7-2=5 the point I1 is I1(7,5).\nline x+y=12 also intersects with line x-y=-8. Lets call this intersection point I2. to find coordinates we have y=12-x and y=x+8 y=y or 12-x=x+8 or 2x=4 or x=2 for x=2 y=12-2=2+8=10 so the point is I2(2,10) \nside I1I2 of the rectangle length is side a. to find length of a we do\na^2=(7-2)^2+(10-5)^2=25+25=50\na=sqrt(50)\nwe need the second side, lets call it b.\nto find length of side b we need at least one more point. lets find intersection of x-y=2 and x+y=2 and call it point I3. y=x-2 and y=2-x or x-2=2-x or 2x=4 or x=2. for x=2 y=2-2=2-2=0 point I3(2,0)\nside b:\nb^2=(7-2)^2+(5-0)^2=50\nb=sqrt(50)\nit means a=b or our rectangle is a square!\nso the area of the square is Area=side^2 since our square&#92;s side is a=b=sqrt(50), the area is 50.\nI hope this helps, I really don't have time to explain it in more details.
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
112219
m = 2\nb = 4\n\nThus, your line's equation is y = 2x + 4\n\nThe x-intercept occurs where the line crosses the x-axis, which is also where y = 0. Substituting 0 for y in the above equation:\n\n0 = 2x + 4\n-4 = 2x\n-2 = x (x-intercept)
238720
So we can find the x-intercept(s)
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)
101438
Y=MX+B\nY=-X ------> Y=-1X+0 \nSLOPE IS -1 AND THE Y-INTERCEPT IS 0
730934
First you have to find the equation of the line that contains these two points (9,1) and (5,5)\nI believe the easiest way to do that is to use slope and y-intercept form (y=mx+b) m= slope and b= y-intercept\n\nto find slope, m = rise/run\nrise = -4\nrun = 4\nrise / run = -1\n\nTo find y-intercept plug in a coordinate pair and use are found slope.( i used 5,5)\ny=mx+b  x= 5 y=5 m=-1 b =?\n5=(-1)(5) +b\n10 = b\n\nSo this is our equation of the two points\ny=-x +10 \n\nNow we can use this equation to find whether  (10,0) is on  the line: Substitute (10,0) and see if the equation is still valid\n\n 0 = (-1)(10) + 10\n 0 = 0 Yes (10,0) is on the same line as the other points
94721
y= 2x + 11 and y=-x + 5 \n\n2x + 11 = -x + 5\n3x + 11 = 5\n3x = -6\nx = -2\n\ny = -(-2) + 5\ny = 2 + 5\ny = 7\n\n(-2, 7)