write the equation in point -slope  form: crossing the x-axis at x=-2 and the y-axis at y=1?
821178
x=-2 ,y=0     &      x=0 ,y=1  \n                     \n\nm=(1-0)/(0+2)\nm=1/2\n\ny=mx + c\n\ny=(1/2)x + c\n\nPut (0,1)\n\n1=(1/2)(0) + c\n\nc= 1\n\nEquation:\n\ny=(1/2)x + 1\n\n2y=x + 2             ANSWER...!!!
610276
points (3,2) and (-3,-10) belong to f and the equation for f is\ny-y1=[(y2-y1)/(x2-x1)]*[x-x1]\nwhere y1=2,y2=-10,x1=3,x2=-3\n\ny-2=[(-10-2)/(-3-3)]*[x-3]\n\ny-2=[-12/-6]{x-3]\n\ny-2=2[x-3]\n\ny=2+2x-6\n\ny=2x-4\n\nso the slope of f is 2\n\nsince g is perpendicular to f the slope is -1/2\n\nand equation would be\n\ny-9=-1/2(x+4)\n\ny=9-[1/2]x-2\n\ny=-(1/2)x+7\n\nso g(x)=-(1/2)+7
35268
slope = (-4-(-6))/(-2-0)\nslope = -1
935919
Please calm down. Now do you want the line between (2,0) and (-2,-2) The way I do it is to get the point vector solution of the line and eliminate t. The point is (2,0). The vector is <-2 - 2, -2 -0>. Just subtract the first coordinate from the second. So the vector is <-4, -2> \nNow set up the point vector equation describing the postion of the graph at time t.\nJust point + vector*t in component form.\nThe x-components: x = 2 - 4t \nThe ys: y = 0 - 2t\n\nThere, that is it. But some nasty professors don't like this form, so we'll eliminate t. We plug y = -2t into first equation and get x = 2 + 2y.\n2y = x - 2. y = x/2 -1.
765225
First put the first equation into slope-intercept form or:\n\ny = mx +b\n\nm = slope,  and b = y-intercept\n\n-3x -3y = 2\n+3x        +3x\n\n-3y = 3x + 2\n/-3     /-3\n\ny = -x + 2/3\n\nSo, your slope for the first equation is:\n\nm = -1\n\nTo have a line parallel to the first line, the slopes must equal.\n\n\n\nUse Point-Slope form.\n\ny - y1 = m ( x - x1)\n\nSince you are using the point (-3, -4)\n\ny1 = -4\n\nx1 = -3\n\nSo your new equation will be:\n\ny - -4 = -1 ( x - -3)\n\n(remember the slope must be the same)\n(a minus a negative = a positive)\ny + 4 = -1(x + 3)\n\ny +4 = -x -3\n   -4         -4\n\ny = -x -7\n\nor you can put it back into standard form ( Ax + By = C)\n\nx + y = -7
876684
y = -x + 6 is in slope intercept form, y = mx + b, where m is the slope.\nSo, you know that the slope of this line is -1. A perpendicular line has the opposite converse slope, so since this slope is -1/1, the slope of the new line will be positive 1/1, or 1.\n\nNow that you have the slope of the new line, you can use that and the point-slope form of a line to find the equation. The general form of a point slope line is y - Y = m(x-X), where m is the slop, Y is the y coordinate of a point on the line, and X is the x coordinate of the same point.\nSo plug in the point and slope to this form and you have:\n\ny - (-4) = 1(x - (-2))\nor\ny + 4 = x + 2\n\nYou can leave the equation in this form, or you can rearrange it so that it is in slope intercept form again.\n\ny = x - 2\n\nI hope this helped!
25053
i don't understand your question too well because it isn't a question but i answer with\nslope intercept form(replace y with F(x)\nmakes it in function notation\nY=mx+b\nm is the slope rise over run\nx is the variable that you can plug in\nb is where the line crosses the y axis\n\nstart with a point on the y axis that is the the y intercept(b)\n\nfrom there do a slope\nexample: y=1/4x+2\n\n\n+++++++++!++++++++++ \n+++++++++!++++++++++ \n+++++++++!+++x++++++ \n+++++++++x++++++++++ start with a point at 2y\n+++++x+++!++++++++++ \n-----------------+ ------------- then move up one and right 4\n+++++++++!++++++++++ \n+++++++++!++++++++++ and a dot down one and left 4\n+++++++++!++++++++++ \n+++++++++!++++++++++ and connect the points, then you would have graphed a segment of the line and just extend that and you solve the problem for your graph
161144
equation of slope is (y2-y1)/(x2-x1)\n\n(8,-2) means x1=8 and y1=-2\n(3,5) means x2=3 and y2=5\n\nso slope= (5-(-2))/(3-8)\n=7/-5\n\nso slope= -7/5
829794
The standard equation for slope-intercept form is y=mx+b. m is the slope and b is the y-coordinate of the vertical (y) intercept\n\n-2x=y+3       Original equation\n\ny=-2x-3        Rewritten in S-I form\n\nm= -2, b= -3\n\nThis means that the graph of the funtion is a line through the y-axis at the point (0, -3) with a slope of (-2)
187139
Yes (-1,0) and (1,-2) share the tangent line y=-x-1\n\nwhich y=-x-1 is tangent to the function y= f(x)=x^4-2x^2-x\n\nNote that (-1,0) and (1,-2) satisfy y=x^4-2x^2-x
561927
Y - Y1 = M(X - X1)\n\nY - 4 = -2(x + 1)\nY - 4 = -2x - 2\n\nY = -2x + 2
554784
x=(1--2)  or (1+2)\ny=(5-3)\n\ny/x=slope\nx=3\ny=2\nslope is 2/3
855184
You don't, really.  y =  mx + b is the slope-intercept line equation, where m = slope and b = y-intercept (the place where x = 0).  I guess you could put both equations in this format, draw the resulting lines, and find where they converge (cross).  That would be easy to do on graph paper, but more difficult to show here.\n\nOne of the other ways to solve the equation system is to substitute one variable for the other.  To do this, you'd first need to put one of the equations in a format where a single variable is by itself on one side of the equation, like this:\ny = 2x - 2\n\nThen plug in the value on the right side of that equation instead of y in the first equation:\nx + (2x + 2) = 4\n3x + 2 = 4\n3x = 2\nx = 2/3\n\nThen plug that back into one of the equations and solve for y\n2/3 + y = 4\ny = 10/3\n\nThe point of convergence for the system is (2/3, 10/3)
371794
slope = (-1-2)/(-1-0) = -3/-1 = 3
698987
Another form of the equation is the point-slope form. If you have two points, find the slope.\n\nm = (a2-b2)/(a1-b1) for points (a1,a2) and (b1,b2)\n\n  Then take one point (a1,a2) and plug it into the following:\n\n(y-a2) = m(x-a1)\n\nIt actually turns out to be the slope intercept form, but if your teacher will accept that form than it is easier since you can just leave it like that.
762224
Your equation is y = 16 (a horizontal line through (0,16).  If you haven't figured out already, it has a slope of zero.\n\nComparing it to y = mx + b, you can see that m = 0.  So the slope of y=16 is 0.\n\nThat means you need another horizontal line that also has a slope of zero, but through the point (-7, -4).  You can probably intuitively figure that it is y = -4, but we can also show that formally:\n\nPutting it into the point-slope form, where m = 0:\n(y - y1) = m(x - x1)\n(y - (-4)) = 0(x - (-7))\ny + 4 = 0\ny = -4\n\nSo either way you get to:\ny = -4
850944
You have 2 points, (x1,y1) and (x2,y2) and you want to find the equation of the line that goes through these points.\n\nThe first thing you want to do is find the slope.\nslope is rise/run so m = (y1-y2)/(x1-x2)\n\nFor slope intercept form , y = mx + b\nwhere m is the slope, and b is the y intercept\nnow that you have found the slope, you can solve for b, the y intercept by plugging in your value for m and one of the given points, either (x1,y1) or (x2,y2).\n\nFor point slope form,  y - K = m(x - H)\nm is the slope, and   (H,K) is a point that the line passes through.\nSince you have already calculated the slope, m, for the slope intercept form, you're already almost done. All you have to do is plug in the point (x1,y1) or (x2,y2). In your question, you are asked for the point slope form of just point (6,2).