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.
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.
Select a set of x values and find the value of y or f(x) - depending on how the parabola is defined. These are the values that you need to graph.
They are all the points where the graph crosses (or touches) the x-axis.
You find the equation of a graph by finding an equation with a graph.
To graph an equation that is not in slope-intercept form, you can use the process of finding points on the graph and plotting them. Choose a few x-values, plug them into the equation to find the corresponding y-values, and plot those points on the graph. Then, connect the points with a smooth line to complete the graph.
A graph is more informative than an equation because a graph is easier to interpret visually, and find all the points and line them up, rather than just a slope which shows no points(data).
So that you can plot out the points of a straight line on graph paper.
In the same coordinate space, i.e. on the same set of axes: -- Graph the first equation. -- Graph the second equation. -- Graph the third equation. . . -- Rinse and repeat for each equation in the system. -- Visually examine the graphs to find the points (2-dimension graph) or lines (3-dimension graph) where all of the individual graphs intersect. Since those points or lines lie on the graph of each individual graph, they are the solution to the entire system of equations.
-- Graph each equation individually. -- Examine the graph to find points where the individual graphs intersect. -- The points where the individual graphs intersect are the solutions of the system of equations.
-- The roots of a quadratic equation are the values of 'x' that make y=0 . -- When you graph a quadratic equation, the graph is a parabola. -- The points on the parabola where y=0 are the points where it crosses the x-axis. -- If it doesn't cross the x-axis, then the roots are complex or pure imaginary, and you can't see them on a graph.
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.
You find out if a problem is linear or exponential by looking at the degree or the highest power; if the degree or the highest power is 1 or 0, the equation is linear. But if the degree is higher than 1 or lower than 0, the equation is exponential.
Draw the graph of the equation. the solution is/are the points where the line cuts the x(horisontal) axis .