1.) Explain how the vertical-line test identifies a graph of a function.?
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1. If you pass a vertical line down the graph, and it only passes through the function once, then ya, it is function.  If it passes through more than once, then it is not a function.\n\n2. Isolate x
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The parent function is y = 6^x. That function contains the point (0,1). Since f(x) = 6^(x+1) is a translation of 6^x one unit to the left, the point (0,1) maps to (-1,1).  \n\nThe graph of f(x) = 6^(x+1) looks like the graph of 6^x translated one unit to the left.
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X & Y is used to find a point on a graph.  X is the number across, and Y is the number up or down.\n\nThat's all I have.\n\nFYI, ur question is a little vague...
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[on] [y=] [(] [x,t,o,n] [+] [2] [)] [/] [x,t,o,n] [enter] [5] [-] [2] [x,t,o,n] [enter] [(-)] [2] [x,t,o,n] [-] [3] [enter] [window] [(-)] [5] [enter] [5] [enter] [1] [enter] [(-)] [5] [enter] [5] [enter] [graph]\n\nYou should have on screen a vertical asymptote at x=0, horizontal asymptote at y=1, a curve going to negative infinity towards 0 on the left, a curve going to positive infinity towards 0 on the right, on the left tangent line y=-2x-3 touches point(-1,-1), and on the right tangent line y=5-2x touches point(1,3).
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Definition:  A function is a one-to-one function if and only if each second element corresponds to one and only one first element. (each x and y value is used only once)\n\nUse the horizontal line test to determine if a function is a one-to-one function.\n\nIf ANY horizontal line intersects your original function in ONLY ONE location, your function will be a one-to-one function and its inverse will also be a function.\n\nTIP:\n\nGraph your function to see if it passes the horizontal line test.\nYou can do the horizontal line test with a pencil.  If the pencil crosses the graph of your function in more than one place (more than one point), then it will not be one-to-one.  If the pencil crosses the graph of your function at only one location, then it will be one-to-one.  \n\nTry it and get back to me.\n\nGuido
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Generally, you break an inequality such as this into two parts and graph each separately.  So you'd be graphing x=-1 and x=3.\n\nThe type of line depends on the sign: if it's a <= or >=, then use solid lines to indicate anything on the line satisfies the original inequality.  If it's just < or >, then use a dashed line.  (This is similar to open and closed circles on a number line.)  Then, you pick points in the different regions that you created with the lines and determine if they satisfy the inequality.  If so, the region gets shaded.\n\nThere is no y dependency, so you basically have two vertical lines for your graph for this problem.  Checking the regions, only the area between the lines satisfies the inequality, so that's the only region that gets shaded.
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A graph is a way to visually show the relationship of different numerical values - there are lots of different types of graphs:  Pie, line, area, bar, etc.\n\nCheck out this site:\n\nhttp://nces.ed.gov/nceskids/createagraph/
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From a table -- see which y-value corresponds to x=0\nFrom a graph -- see where the graph crosses the y-axis\nFrom an equation -- get the equation in y=mx+b form; the y-interecept is b
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This is true (but why didn't you put it in the math section?).\n\nA rational function has a variable in the denominator, like 1/x.  When the denominator's value is equal to 0, one of two things happens--the function has a vertical asymptote, or it has a "hole" in the graph.\n\nWhat usually creates the hole is when you have a value in the top and bottom that "cancels" out, leaving a different rational function to graph.  For example, what does the graph of f(x) = x/x look like?  Since x/x = 1 for any value EXCEPT 0, the graph should (basically) look like the graph of f(x) = 1.  However, the fact that f(x) = x/x is undefined when x = 0 leaves a hole in the graph at that point.
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Hi Bailey,\n\nI'm really glad you're still trying and haven't given up!\n\nFirst I have to confess I'm slightly confused by the wording of your question. WHAT STARTS AT ZERO AND GOES TO SIX? The X-axes, Y-axis or the line? Do you have the co-ordinates of 2 points on the line? \nHave you been given values of x ranging from 1 to 6 and from that you have to calculate the intercept? PLEASE CLARFY BY ADDING DETAILS\n\nNow that you are an expert at graph sketching (or at least understand them better now) you will know that both the X- axis and Y-axis extends infinitely in both positive and negative directions.\n\nPLEASE REMEMBER THAT THE X AXIS GOES ACROSS (horizontally ie left to right) and THE Y AXIS GOES UP (vertically, up and down). \n\nNow try to follow this...\nThe intercept is the value of y (point on the Y vertical axis) where the line 'cuts' it. At this point X always must equal zero (by the definition of a Euclidian plane, but don't worry about it).\n\nTherefore put the value of X=0 into the formula and you will get:\ny= 0m +b \n= y=b \nso the intercept point has coordinate (0,b) ie X=0 and Y+b\n\nIf there was no value on the line when X=0 then the line DOES NOT not intercept the Y axis therefore b would not exist for this line. Such a line occurs for verticle lines (straight up) such as x=1,x=2,x=3,x=4,x=5,x=6…to infinity.\n\nIncidentally if you were given 2 points and asked to find the intercept you should use ALGEBRA. Compute the gradient (y2-y1)/(x2-x1). Then compute the intercept as the balancing value to make the equation work. \nFor example (5,6) and (7,10). \nM= 10-6/7-5=4/2=2. \ntherefore y=2x+b. \nChoose one of the points and substitute the values.  \nSay 6= 2(5)+b so b= 6-10 = -4.\n\nDoes this answer you query?
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what is on the axes?\n\nthis is the best i could find for you...\nIf a graph shows time on the horizontal axis and speed on the vertical\naxis, a straight horizontal line across the graph would indicate?\n\na. no motion\nb. constant speed\nc. steady acceleration\nd. maximum speed\n\nANSWER:\nb. constant speed.\n\n\nIn effect, the horizontal line in the graph indicates that for every\nvalue of the horizontal axis (time) corresponds the same value on the\nvertical axis (speed), i.e. the same speed for every time, that is a\nconstant speed.
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well, you can pick an X and plot a Y (meaning, take values close to the Origin for X and put them into the equation (and inequality), and see what Y you get for each of them\n\nfor instance, if you pick 1 for X, and put it into the equation, you get y = 1. . so now you have one pair (one point you can already plot on your graph, the point (1, 1).\n\nNow pick 0, and you get a Y of 0, so you have the point ( 0, 0), so now you have two points.  \n\nnow you need to see what it does in negative territory.\n\nPick some negatives (like -1 ) and you get 1, so you know that the point ( -1, 1) is on the graph. . \n\nif you pick positive 2 for x, you get 2 for y. . so ( 2, 2) is on the graph. . \n\nif you pick -2 for X, you get positive 2 for Y, so ( -2, 2) is on the graph.\n\nnow you can plot the graph of that line. . it's two lines, perpendicular to each other that meet (and change direction) at the Origin (the point (0, 0).\n\nAs for the inequality, just graph it as though it is an equals sign instead of an inequality sign, and then then you know where the line is, and you find out which points are on which side of that line by picking X's and finding what Y value they give you (when you put them into the equation (or into the inequality).\n\nhope this helps, \n\ngood luck and God bless you.
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It should be a function because one x input value gives only one y output value. Try the vertical-line test. But the function does not have an inverse though.
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You restrict the domain of the function as necessary in order that it pass both the vertical and horizontal line test.  Once you do that, f^-1 exists.  For example, if f(x) = sinx, you can restrict its domain to (-pi/2, pi/2). This function will then pass both h & v line test, and its inverse will then exist.\n\nanother example - f(x) = x^2.  A parabola opening upward whose vertex is the origin, and fails the h-line test.  If you restrict domain to (>=0),  f^-1 will then exist.
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Vertical asymptotes will occur where the function is undefined. Usually this happens when you have a division by 0. In this case, when x = 0 you can't evaluate the function, and as it turns out, there is a vertical asymptote there.\n\nHorizontal asymptotes are found by taking the limit of the function as x goes to either positive or negative infinity. In this case, when x gets really big, the function gets really close to 1. As it turns out that happens for both negative and positive infinity.\n\nSo the answer is vertical asymptote at x = 0 and horizontal asymptote at y = 1.
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Ok, since you didnt give any equation, i would assume that the graphs are linear in nature\n\nThen for find the points where the graph intersects the Axes,\n\n  Substitute x=0 in the equation to find the point where the point where the graph intersects the Y axis\n\n    Substitute y=0 in the equation to find the point where the point where the graph intersects the X axis\n\nEg.   2x + y = 4\n\nSubstituting y=0,\n\n                          2x=4\n                           x=2\n\nHence the the graph meets the x axis at (x,y) that is (2,0)\n\nSimilarly, substituting for x=0, we get..\n\n                         y=4\n\nTherefore the point where hte graph meets the x axis is (x,y), in this case (0.4)
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Asymptotes are linear areas of the graph where no values will satisfy the equation - in other words, the value of the function approaches either 0 or +_infinity.\n\nObviously, x can never be 0, so x = 0 would be a vertical asymptote. But for any non-zero value for x, y can be just about anything. However, from the function, you can see that the numerator is a constant, so y will never reach 0, no matter how large x gets. So y = 0 is a horizontal asymptote.