The graph of y=2x is a straight line passing through the coordinate point (-3/2, 3/2).  Is this true or false?
914486
just plug -1.5 in for x and see if it equals 1.5.       \n2 X -1.5 = -3                  -3 does not equal 1.5\nso the answer is false
733792
Yes it is a function because by the verticle line test it only intersects that line at 1 point.  \n\n\n It is a pretty simple test.  if a verticle line hits the graph at more than 1 point it is not a function.
921939
Good Evening\n\nI think we've been avoiding this question because we can't graph here... however to get you started, create an x|y chart for each line, plug in a couple of values, and connect the dots  (for a line you only need two points, but some books recommend finding three, so you can verify you didn't make a mistake somewhere)\n\nx+y = 8\nx|y\n----\n4|4\n3|5\n2|6\n\nx - y = -4\nx|y\n0|4\n1|5\n2|6\n\nGraph the two lines, and they should intersect at (2,6) since it appears in both of the charts...\n\n~i~
608711
i think it's a parabola.. im not sure. u should try plotting it on a graph.. substitute  the x and y values and see what form comes out..
620528
There are two equations there, not one.\n\nFor the first equation, plug the numbers on your X-axis into "x", calculate "y", then draw the points on your graph.  I'll get you started:\n\nx=0, y = -3\nx=1, y = -1\nx=2, y = 1\netc.\n\nThe second equation simply means that x = 1.5 and y = 3.  This is just a single dot on a graph.
942634
If the fucntion y intersects a straight line in 4 distinct points, say x1, x2, x3,  x4 then its derivative y' has 3 roots r1, r2 and r3 such that x1 < r1 < x2 < r2 < x3 < r3 < x4. This means the graph of y' intersects the x-axis (which is a straight line) in 3 distinct points. Therefore its derivative, i.e. y'' has 2 distinct roots, say s1 and s2 with r1 < s1 < r2 < s2 < r3.  But y'' = 4.3x^2 + 3.2.cx + 12.2 = 6( 2x^2 + cx +4), and y'' has 2 distinct roots iff discriminant =c^2 - 2.4 =c^2 - 8 > 0, i.e. iff c runs in (-infinity, -2sqrt(2))union(2sqrt(2), + infinity).\n\nI hope this will help!
473627
on a graph every point has an x and y coordinate.  The vertical(up-down) is the Y-coordinate, the horizontal (left-right) is the x-coordinate
149070
When x=0, that line will cross the y-axis. When y=0, that line will cross the x-axis.
796114
Graph equations are of the form y=mx+c\n\nm is the slope or gradient\nc is the y intercept.\n\nSo, for y=-x+1, \nm= -1\nc = 1\n\nAnd for 2y=3x, you have to rearrange it to make it look like y=mx+c.\n\nSo divide by 2.\n\ny=3x/2\n\nm= 3/2 \nc =0\n\nHope you understood my explanation :)
896230
the y coordinates are the same i.e. -1, \nso the line is parallel to the x axis and passes through -1 on the y axis.\nits slope is parallel to the x axis.
1134068
No, \nfor f(x)=(x+5)/(x-9), you substitute 1 for x, then solve for f(x) to obtain the y-value at x=1. So:\n\n                                 f(1)=((1)+5)/((1)-9)= -6/8\n\n                               which shows that at x=1, y= -6/8 = f(1)\n\nThe point (1, -5/9) is not on the graph.
1175454
y=-3x\nand\ny=3x\n\nare perpendicular, but only if you change the scale into 1/3 of the y-scale(meaning, you divide the y-scale by 3). Otherwise y = -3x and y = 3x are not perpendicular.\n\n^_^
974622
The graph of y = 1/x^2 can be graphed on a calculator and has even symmetry with the left half looking a lot like the left half of the hyperbola y = 1/x except it cuts upward more sharply.\n\nx = 1/y^2 is just this function's inverse which is found by exchanging the positive x and y axes and the negative x and y axes. \n\nImagine that you have a transparency with the graph on it with your right hand holding the top of the sheet and your left hand holding the bottom of the sheet. You flip the page over in such a way that your right hand is on the right side of the sheet and your left hand is on the left side of the sheet.\n\nSince the original graph has even symmetry, this process is also equivalent to rotating the graph of y = 1/x^2 clockwise by 90 degrees.\n\nSo it's the same kind of curved triangular pattern except about the x-axis rather than the y-axis.
763572
Hi!\n\nSo, what you are looking at is a linear equation.  You can tell this because there is only one instance of the variable x, and it is not raised to any power.  So, the graph will take the form of a line.\n\nAll linear equations take the form f(x) = mx + b\n\nWhere m is typically called the "slope" and b is called the "y-intercept".   The y-intercept is the point on the graph where the function intercepts the y-axis (the one that goes from top to bottom on the graph).  So, right away you can tell that the y-intercept in your equation is -5 (negative 5).  What this means is that the graph intercepts the y-axis at the -5, or in other words, there is a point on the graph at the location (0,-5).  You can put a dot there.\n\nThe next thing you need to see is the slope.  In you equation, the slope is -2 (negative 2).   Slope really means "rise over the run", or in other words, how much the graph rises for every point it goes to the right.  A slope of negative 2 means that for every increase in x of one unit, the y value decreases by negative 2.   \n\nSo, starting at the point (0,-5) which we determined above, go down 2 units and to the right one unit (putting you at the point (1,-7).  That's another point.  You can do it again and find the next point which is (2,-9) and put another point.  You now can put your ruler on the graph and draw the rest of it.\n\nIt's good to understand how this type of equation works.  You could have also made a table of x values and y values, and simply computed a series of y values for a series of x values (that you picked).  Then you could have drawn the graph also.  But understanding this type of equation will help you in the future.
871684
The graph of this equation looks like a V. It's the same as the graph of y=2x, except that the part which should bein quad III is actually in quad II.\nFor this graph, the vertex is when y=0. \nSo the vertex is at (0,0).
1124994
you can use the y=mx+b formula for a line.  \n\ny is the y-coordinate of a point\n\nm is the slope, x is the x-coordinate of a point\n\nb is the y-intercept (where the line touches or intercepts the y-axes).  \n\nsince you know the y intercept of the line (passes through origin, so it's 0), the value of b would be 0 and  you also know the slope of the line, which is two, the equation for the line would look like this.\n\ny=2x + 0 or y=2x\n\na point on the line would have coordinates that would fit the equation ---- y has to be two times x; x has to be 1/2 of y.\n\nhere are some points on this line: (0,0), (1,2), (-1,-2), (2,4), (-2,-4).\n\nhope that helped :)
318213
eh? you using simultaneous equations with a straight line equation?\n\ny = mx + c is a straight line equation, where m is the gradient & c is where the line cuts the y axis.\n\nThere for lets say y = -2x + 8:\ny = mx + c\nCuts x-axis Put y = 0\n-2x + 8 = 0\n-2x = -8\nx = -8/-2\nx = 4\n\nHope that was correct :| lol