Since there are no "following" equations, the answer is NONE OF THEM.
Since there are no "following" equations, the answer is NONE OF THEM.
Since there are no "following" equations, the answer is NONE OF THEM.
Since there are no "following" equations, the answer is NONE OF THEM.
the left end of the graph is going in a positive direction and the right end is going in a negative direction.
base
You can move it up or down by adding a constant, call it c. Let c>0 Y=radical(x)+c move it up c and y= radical(x)-c moves it down c. You can move it to the right by subtracting c inside the radical sign. Let c>0 y=radical (x-c) moves it to the right c units. y=radical (x+c) moves it to the left c units.
FALSE
Before you start with limits, you should know that they are quite similar to finding the instantaneous rate of change. The limit of any given point (a) on the graph of a function would be the value the graph converges to at that point. The limit, in other words, is the slope of the tangent at a certain point on the graph. For example, take the graph of y = x [Which is the same as f(x) = x] Now, when you graph that function you get a perfectly diagonal line. You can just start at the point (0,0) on the graph and then for each point, go up 1, right 1. Do the same for the left part of the graph, going down 1 and left 1. Now that you got the graph, take ANY value of x. Say you take 5. Now what point is your FUNCTION approaching from EACH side. So its clear that your function is approaching a value of 5 on the y-axis when x=5, from each side i.e. the graph approaches 5 on the y-axis from the left and the right when x =5. Remember that for a limit to exist, the graph should always approach a certain point from BOTH directions, left and right. Consider the graph of y=x2. At x =5, y = 25. Now since the graph approaches the point 25, when x = 5 from both left and right sides, the limit as the graph approaches x=5 is 25!! Remember that it does NOT matter if the graph is defined at the point at which you are finding if the limit exists, what only matters is if the graph is approaching the point from both sides. So to say, you can have a hole at (5,25) and still have the limit as 25. Now there's a specific way of writing limits. Have a look at this image: http://upload.wikimedia.org/math/e/8/7/e879d1b2b7a9e19d16438c24fb8a7990.png Okay, I'll describe what the image states. All its saying is that as x approaches point 'p' on the function f(x), the limit is L. So, to say for the example I just did above, you have have '5' instead of 'p', and 'L' would be replaced by '25'. Now, say the limit at x=2, for the function f(x) is 10, but you actually have a hole at the point (2,10). And you have a DEFINED point at (2,12). IF your graph is still approaching the hole at (2,10) from both sides, then your limit will still exist. Moving on, suppose a point is x = 3 on a certain graph. So, in 'calculus terms', when the graph is approaching 3 from the left side it would be written like 3- while approaching from the right would be 3+.
Which transformations could have been used to move the platter to the new location? A. a translation 9 units left and a translation 3 units down B. a reflection across MN and a translation 4 units left C. a reflection across MN and a translation 8 units left D. a rotation 180° clockwise about N and a translation 4 units left
Go what equals 5 units to the left
On all of the graphing calculators, they have the graphing things on it. First to make a graph, you need an equation. when you have the equation ready, you just press the button on the top left (for the most calculaters), the "y=" button. Then you will get a lot of empty lines for you to put the equations on. After you type all the equations, you will have to press the "graph button on the top right. then you will get a graph.
it is the same as a sin function only shifted to the left pi/2 units
(x,y)--(x-4,y+6)
Translation is moving a graph to the left or right, up or down (or both). Given a quadratic equation of the form y = ax^2 + bx + c, if you substitute u = x - p and v = y - q then the graph of v against u will be the same as the x-y graph, translated to the left by p and downwards by q.
If y = f(x), then y = f(x + c) is the same graph shifted c units to the left (or right if c is negative) along the x-axis For y = x, by changing x to x + c, the above shift is indistinguishable from shifting the graph c units up (or down if c is negative) the y-axis.
(x1, y1) = (x - 8, y + 9)
If the question is what I think it is, then it means a translation of the graph by 1 unit to the left.
translation
Press APPS and select Inequalz to turn on inequality graphing. (If it is already on, choose continue.) Press Y= and there will be a little X= in the upper left-hand corner of the screen. Select it and press ENTER. The equations will change to X=. This allows you to graph vertical lines (you cannot graph equations like x = y, it must be x = a constant). Put 1 into one of the X= equations and press graph. (To turn of inequality graphing, press APPS and select Inequalz. Choose Quit Inequal.)
Yes. For example, if you want to shift the graph 5 units to the right, you must replace every instance of "x" by "x-5".