Find points where y=0?
954093
k1ax - k2x^2 = 0\nx(k1a - k2x) = 0\n\nat x = 0 and...\nk1a - k2x = 0\nx = k1a/k2\n\nso at x = 0 and a*k1/k2
399118
y-int= (0,-5)\nx-int= (-1.2,0)
316569
y = mx + c\n(0) = (2)(-1) + c\nc = 2\n\ntherefore y = 2x + 2
19244
(Y-y1) = m(X-x1)\non point (4,7) => Y-7 = m(X-4)  (I)\non point (2,5) => Y-5 = m(X-2)  (II)\n\n(I&II) ==> m = 1
805734
Point A: (1,4)\nPoint B: (2,1)\n\nTake the coordinates one at a time.  From A to B, the X coordinate goes from 1 to 2, a difference of 1.  Twice the existing difference of 1 is 2, which we add to Point B's X coordinate, 2, to get 4.\n\nDo the same thing with the Y coordinates.  From A to B, the Y coordinate goes from 4 to 1, a difference of -3.  Twice the existing difference is -6, which we add to Point B's Y coordinate, 1, to get -5.\n\nCombining the X and Y coordinates, we get the final answer, (4,-5).
133024
y = 0.\n\nETA: That's if we're to assume that the value of x is constant between the equations.
826422
Find the slope of the line that connects those points.
389930
Write the equation in the point-slope form, then substitute 0 for x and solve for y.  That is the y-intercept.
679397
y = 0.5x-4 \n\nor similarly\n\nx= 2y +8
1119524
well using the y = function, you may need to create a positive and negative graph\n\nfor a circle with a center of 0, 0:\n\nsquare root ( x^2 - r^2)  (replace r with your desired radius)\nand\n- square root( x^2 - r^2)
444240
-y=2-x\n\ny=x-2\n\nslope is 1\ny intercept is (0,-2)
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
754976
It's true.\nThe answer is indeed y = 0.\nYou can find this by dividing each side by -4.\n-4y = 0\n-4y/-4 = 0/-4\ny = 0\nRemember that zero divided by any nonzero number is always zero.
55111
The formula for a midpoint is (X1+X2)/2 for X and (Y1+Y2)/2 for Y.  For your specific problem, the answer would be: (6,9)
374950
Plug in x, solve for y as follows:\nIf x=0 then solve for y and get y= 3/2\nif x=10 y= -10/2+ 3/2 = -7/2\nCreate a graph using the horizontal as the x scale and the vertical as y.  Plot the above and you will see that it's a straight line.
653036
when you have 2 equations and you have to solve for a point(s) those points are the intersections of thir plots.\n\nSince their are intersections, y from first eq has to be equal to y from second equation for one or more common x values.\n\nKnowing this you can do y1 - y2 =0\n\nIn this porblem you have:\ny1=e^(-x)+1\ny2=ln(x)+3\n\nso \n\n( e^(-x)+1 ) - ( ln(x)+3 ) = 0 \n\nequals to:\n\ne^(-x)-ln(x)-2=0\n\nNow any x that satisfy that eq is a intersection point, therefore an answer of your problem.\n\nFor this eq (and many others) Newton method for finding zeros (or roots) is the best option. (check link on source)\n\nLooking at plots, this curves has only one interception point, close to x=0.3\n\nBy aplying this algorithm, starting with a value of x0=0.5 you will find:\nx1=0.231...\nx2=0.281...\nx3=0.286...\nx4=0.286713167526\nx5=0.286713169911\n\nyou can aproach as much as you need.\n\nFinally, to find y1 value (or y2 because are the same) \npick one of the ecuations and solve for the x value you\nfind.\n\ny2=ln(x)+3\n\ny2=ln(0.286713169911) +3\ny2=1.75072702906073\n\nthen (x,y)=(0.286713169911 , 1.75072702906073)
147089
y intercept: (0,4)\nx intercept: (3,0)