graph line through the point (5,-3) with undefined slope?
953419
The slope intercept form of the line is \n\ny = mx + b where m = slope and b = y-intercept\n\nSince (5, -3) lies on the line then\n\n-3 = 5m + b\n\nie 5m + b - 3 = o\n\nSo any line in which m = (3 - b)/5\n\neg b = -2, m = 1 So y = x - 2\nb = 8, m = -1 So y = 8 - x ....... and so forth\n\nGraphing m against b would give all the values of m, b that satisfy the given condition.
532403
The slope of y = (1/3)x - 5 is m = 1/3\nA line that is perpendicular to the above line will have slope m = -3.\nIf two lines are perpendicular, then the product of their slopes is -1.  For this case, we have (1/3)(-3) = -1, so -3 must be the slope.\nSince you're not given a point, any equation of the form \ny = -3x+b will work.\nJust pick whatever value of b that you want.\nSome examples, \ny = -3x\ny = -3x + 5\ny = -3x - 7
835994
The slope of any line is dy/dx. The x axis has no dy; i.e., dy=0 so dy/dx=0. The y axis has no dx; dx=0 so dy/dx = undefined.
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.
971807
(0,0)\n(3,0)\ndraw the graphs and then find them out!\nif u can't, then mail me!
861221
looks like the vertex is (-5, 3)
155791
Here is a good page describing slope.\n\nhttp://mathforum.org/cgraph/cslope/slope.html
560706
You need to get the y alone on one side of the equation. Subtract 5x from both sides and you get:\n-3y=-5x-3\nDivide each variable by -3 to get:\ny=(5/3)x+1\nNow you can graph using this equation.\nThe y-intercept is 1 and the slope is 5/3.\nHope this helps.
869140
Just graph it like an equality, and then your answer is everything either above (>) or below (<) the line.  For example, in the last inequality, the answer is everything above the line y = -2, not including the line itself.
816764
This is equivalent to y < -3.  Since y = -3 would be a HORIZONTAL (not vertical) line, you'd graph a dotted line going straight across at -3 on the y-axis.  Since you want the y-values that are LESS than that, you'd shade below that line.
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
985391
Use point-slope form:\nFirst, calculate the slope: (-2 - 0) / (5 - 0) = -2 / 5.\nHere is the general point-slope form:\n(y-y1) = m*(x-x1)\n(y+2) = (-2/5)*(x-5)\ny + 2 = -2/5*x + 2\ny = -2/5*x\n\n______\n\n\nFor parametric form, there's not that much to it:\nx = 0 + 5t\ny = 0 + -2t\n\nFor clarity, here's an example with another set of points:\n(2,-3) and (4,8)\nx = 2 + 2t\ny = -3 + 11t
97021
Using the points you can work out the slope. That is 0.\nThen use one set of points and the slope to work out the equation of a line formula which is:   \n         y-y1=m(x-x1)\nm is the slope \n       y+4=0(x-11)\n       y+4=0\n       y=-4
907088
The slope is 1/3 and the point is (-2, -4).
826422
Find the slope of the line that connects those points.
734729
problem1\n\nsince you have said P (4,-5) is a point on the function, then the corresponding point will be (4,-5) .. :)\n\nproblem2\n\nfor a line having y intercept 3 will have following form\n\ny =mx+3\n\nwhere m is the slope of the line. Now it is said to be perpendicular to y =2/3x-4. So slope for this is 2/3\n\nsince slopes of two lines perpendicular to each other if multiplied yields -1\n\nm X 2/3 = -1\n\nor m = -3/2\n\nso the required line will be\n\ny = -3/2 x+3\n\nproblem3\n\nf(x) = 1/x^2 -2x-8\n\nto make the f(x0<0, let us try with some arbitary values of x\n\nif x = 1, f(x) = -9\nif x=2, f(x) = -11.75\nif x = 0, f(x) = infinite (due to 1/x^2)\nif x = -1, f(x) = -5\nif x= -2, f(x)= -3.75\nif x=-3, f(x) = -1.88889\nif x=-4, f(x) = 0.0625\n\nHope this helps
71640
you find to points on the line \nexample (2,5) and (1,6)\naight m= slope\nput the points in this simple equation\ny2-y1 (over) \nx2-x1\n\n6-5 over\n1-2\nwhich Equals  \n M= 1/ -1\nHoped it helped\nor try asking the teacher for help after school or at lunch