I assume you're fitting an relation of the form y = aebx.
Take the natural logarithm of both sides of the relation: ln y = a + bx (1).
Suppose your points are (-1, 7.39) and (1,0.14). Because we have a relation between the logarithms of the y co-ordinates and the x co-ordinates themselves let us replaces the y co-ordinates with their logarithms to obtain the points (-1, 2) and (1, 2). Now we can use this points to obtain two simulaneous equations in the transformed equation (1) and solve it as usual to obtain a = 1 and b = -2.
The fitted relation is y = e-2x.
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They are all the points where the graph crosses (or touches) the x-axis.
From the equation, the y intercept is simply determined by setting x = 0. The x intercept(s) are generally much harder to find: you will need to find the solutions of y = 0 [or f(x) = 0]. From the graph the intercepts are the coordinates of the points at which the graph crosses the axes.
1. Decide if the graph looks like any standard type of graph you've seen before. Is it a type of sine or cosine? A quadratic? A circle or ellipse? A line? An exponential? (You get the idea.) If you can't find a standard type to match your desired graph, pick one that looks close to it and recognize that you will be doing an approximation to your function.2. Once you have an idea of what you're graph should be like, think about the equations that are used to describe that graph. Where do the numbers go and how do they affect how the graph looks/moves/ behaves? Some functions, such as circles, hyperbolas, and quadratics, have standard equations with variables based on the important features of the graph (such as the center, maximums or minimums).3. Find the important and/or interesting parts of the graph and use them in the equation. As stated before, ellipses and such have special equations to describe them. Sines and cosines require the amplitude, frequency, and phase shift.4. Check your equation if you can. It's always good to plug a few of the points that are in your graph to make sure your equation is accurate. It's especially good to try out points you did NOT use to find your equation. If it works for these, then you probably did it right.
For 5y-15=5x 5y=5x+15 y=x+3 The y-intercept is (0,3) To find the x-intercept, substitute 0 in for y in the above 0=x+3 x=-3 So the x-intercept will occur at x=-3, making it (-3,0) Mark these two points and draw a straight line along the two points and you will get your equation's graph.
Graph the equation then find the x intercepts.