If you can find a function f which corresponds to the graph or table, then you can simply evaluate f(0) to find the initial value of y. If there is no function, then looking at the graph, you should find the point where the graph intersects the y-axis. In layperson terms, the point where the graph and y-axis cross each other, that is the initial value of y.
TI-84: How can I trace a Y value?
Unfortunately, that's not possible on the TI-84 Plus. The graph works by calculating a y-value for each x-value, and so when you use [2nd][Trace](Calc) > 1:value, it's merely plugging the value of x you input into the Y-variable, and spitting out the corresponding y-value. If the function's inverse is still a function, then you could merely graph the inverse and trace x (which would be the same thing as tracing y).
Finding function values on a graph!?
The x in the function defintions f(x) and g(x) is simply where you plug in your numbers -3 and 0. On the graph the x-axis is the horizontal line where you set the x values. So to find the value of f(g(-3)) find -3 on the x-axis then go up or down until you intersect the line g(x) and follow that point back horizontally until you intersect the y-axis. That is the value of g(-3). Now find that value on the x-axis then go up or down until you intersect the line f(x) and follow that point back horizontally until you intersect the y-axis. That is the value of f(g(-3)). Repeat the same process for x = 0.
How do i find a function when given a graph?
If you are given the graph then you know what type of function it represents. eg If its a straight line then the form of its equation is y = mx + c The y-intercept (c) can be read off the graph. The gradient (m) can be found by choosing any two points on the line then evaluating Δy / Δx (ie change in y-coordinate ÷ change in x-coordinate) The value for c and m can then be substituted into y = mx + c and you have your equation. If the graph is a parabola, then it represents a quadratic function and the form of its equation will be y = ax^2 + bx + c Use the graph to read off the y-intercept ( ie when x = 0) this will give you the value of c. Use the graph to read off the coordinates of the x-intercepts (ie when y = 0). Sustitute the values of c and of one pair of x and y values into y = ax^2 + bx + c to get an equation with an a and a b to solve for. Do the same thing but using the cordinates of the second point as the values of x and y to get a second equation with an a and a b to solve for. Solve these two equations simultaneously to find a and b Then substitute the values of a, b and c into y = ax^2 + bx + c and you then have the equation of the quadratic represented by the graph. Note if the graph doesn't have any x-intercepts or if there is only one, just choose the coordinates of another convenient point whose coordinates are easy to read off the graph. The equation of a cubic function can be found in a similar way but three equations will be need to find the a, b, and c in the equation y = ax^3 + bx^2 + cx + d and so on The starting point is always to write down the general form of whatever function is represented by the graph. Then find the coefficients and the constant for the particular function so that its equation can be written.
Calculus, max/min values of trig functions?
The graph is always increasing except where the derivative is zero. f'(x) = 1 - sinx, which is zero when sin x = 1, so x = pi/2, 5pi/2, 9pi/2 etc to find local maxima and minima, take the second derivative at the turning point f"(x) = -cos x at pi/2, cos x = 0 This tells you that there are no local maxima and minima. What you have is a point of inflection at pi/2 + 2npi Think about it like this: If you have a function that is increasing everywhere, you can't have local maxima or minima. By definition the gradient has to change sign either side of the turning point. f'(x) = 1 - sinx is always >=0. If the gradient is never negative, you can't have a max or min anywhere.
Excel Graph, line intersections?
You could also use "Solver" to find the value at an intersecting point. What worked for me was this: when graphing two lines and looking for the intersection of the lines, I had Excel create two regression lines of my data points (called "Linear Trendline" under "Trendline" drop down menu of "Chart Tools" navigation). Then I set the two equations equal to each other like this: =((1*M27)+ 2)-((2*M27)+1). This example would be for two equations: y=1x + 2, and y=2x + 1. (make sure to use the same empty cell/ M27, for X of each equation.). After that, go to the Data tab and click on "Solver". For the "Set Objective," reference the cell that you used to set the equations equal to each other. For the "By Changing Variable Cells:" type in the cell you referenced as the X value (M27 in the example). Click Solve. Click OK to keep solver solution, or reset it to original values so you can do other things. It will give you the X value of the interception of your lines. Plug X into one equation (or both if you want to double check accuracy of value) to get your Y value. And you're all set! If you're looking for more help on Excel, I've gotten a lot out of Mike Girvin's ExcelIsFun videos on YouTube. He's made literally thousands of Excel videos and really makes it fun to learn about in a logical way. Here's the link to his page: https://www.youtube.com/user/ExcelIsFun/about (I'd start with the Excel Basics series). Okay, good luck and cheers from MN! Nick
Suppose you have a function y = f(x).To find the y-intercept of a function, plug in x = 0 into the function. Thus, the y-intercept will be f(0). For a function, there can be at most ONE y-intercept. This is because of the onto property (a.k.a. the vertical line test), which assures that every function has, at most, one y-intercept.To find the x-intercept(s) of a function, solve for every x-value that satisfies the equation f(x) = 0. In other words, to find the x-intercept(s) of a function let y = 0. There can be multiple (or infinitely many) x-intercepts, only one x-intercept (such as in a linear function) or no x-intercept.If you wish to find the x-intercept(s) of a multi-variable function such as z = f(x, y), plug in y =0 and z = 0 and solve for x. To find the y-intercept, plug in x = 0 and z = 0 and solve for y.As a general principle, if you want to find the x-intercept of the function w = f(x, y, z, a, b, c…), plug in zero for every variable except for x and solve for x. If you wish to do the same with the y-intercept, plug in zero for every variable except for y and solve for y.I hope this helps!
Your worksheet layout may look great for filling out with pencil and paper—but it makes the desired chart extremely hard to build.To make the chart easy to build, put your dates in one column, times in another column and counts in a third column. In a fourth column, you can concatenate the date & time using a formula like:=A2+B2Both dates and times are stored as numbers, with each passing day increasing the stored value by one, and each passing hour increasing it by 1/24. That’s why you add them to perform the concatenation.If you might want to look at the values for only the mid-morning times, I suggest using a PivotChart. You create one by selecting a cell in your data, then using the Insert…PivotChart…PivotChart menu item. This will eventually take you to a somewhat unfriendly looking PivotChart taskpane. Drag the Date from PivotTable Fields to the Axis (Categories) box, the Time to the Filters box, and the Value to the Values box. In the filter at the top of the resulting PivotChart, you may choose which time you are interested in.
How do you find maximum r values of polar equations?
1 . r = 5 since r is constant the maximum value of r = 5 2. r = 3 sinθ here r value depend on the value sinθ r will be maximum when sinθ is maximum sinθ is maximum when θ = 2k pi + pi/2 where k = 0,1,2,3.... when θ = θ = 2k pi + pi/2 where k = 0,1,2,3.... sin θ = 1 then r = 3 sinθ = 3 * 1 = 3 therefore the maximum value of r = 3 3. r = 4 ( 1 + sin θ ) here also r depends on sin θ value , r will be maximum when sin θ is maximum sin θ will be maximum when θ = 2k pi + pi/2 where k = 0,1,2,3.... => sin θ = 1 then r = 4 ( 1 + 1 ) = 8 therefore maximum value of r = 8
What you are really asking for is not to read data from the chart, but to interpolate values from the data you originally put into it. In the data you provided there simply is no data pair (x,50), so your chart will never be able to return x without doing some kind of interpolation between the two surrounding points (2.5,55) and (5,32).Of course Excel shows lines between the points in your chart, and that suggest that there is data, but there really isn't. The lines are simply a linear or some kind of polynomal interpolation between the surrounding points.So the only right way to do this is by adding a trend line.Select your data points and choose insert trend line. It is up to you to decide whether it will be a linear trend line, a logarithmic or a 3rd order polynomal. Make sure you check the "show equation" box. Then Excel will add a trendline and show its equation. Using that equation you can simply calculate X from any given Y. As far as I know the only way to do this is type the equation manually in your workbook. Keep in mind however that any trend line is always an approximation, and that you better not use it to extrapolate.Below two screenshots (in Dutch).