Then right click on the curve and choose "Add trendline" Choose "Polynomial" and "Order 2". Enter the points in cells as shown, and get Excel to graph it using "X-Y scatter plot". You could use MS Excel to find the equation. This Wolfram|Alpha search gives the answer to my last example. We just substitute as before into the vertex form of our quadratic function. We can see the vertex is at (-2, 1) and the y-intercept is at (0, 2). Here's an example where there is no x-intercept. Note: We could also make use of the fact that the x-value of the vertex of the parabola y = ax 2 + bx + c is given by:
So our quadratic function for this example is In this example, the blue curve passes through (0, 1) on the y-axis, so we can simply substitute x = 0, y = 1 into y = a( x − 1) 2 as follows: What is the value of " a"?īut as in the previous case, we have an infinite number of parabolas passing through (1, 0). If we use y = a( x − h) 2 + k, we can see from the graph that h = 1 and k = 0. This parabola touches the x-axis at (1, 0) only. The next example shows how we can use the Vertex Method to find our quadratic function. (If there are no other "nice" points where we can see the graph passing through, then we would have to use our estimate.) It's near (−0.5, −3.4), but "near" will not give us a correct answer. In our example above, we can't really tell where the vertex is. We can write a parabola in "vertex form" as follows:įor this parabola, the vertex is at ( h, k). Vertex methodĪnother way of going about this is to observe the vertex (the "pointy end") of the parabola. We note that the " a" value is positive, resulting in a "legs up" orientation, as expected. So the correct quadratic function for the blue graph is Substituting a = 1.5 into a + b = 3, we get b = 1.5. Substituting c = −3 in the first line gives:Ĥ a − 2 b = 3 and substituting into the second line gives: Using our general form of the quadratic, y = ax 2 + bx + c, we substitute the known values for x and y to obtain: On the original blue curve, we can see that it passes through the point (0, −3) on the y-axis. We can then form 3 equations in 3 unknowns and solve them to get the required result. To find the unique quadratic function for our blue parabola, we need to use 3 points on the curve. So how do we find the correct quadratic function for our original question (the one in blue)? It turns out there are an infinite number of parabolas passing through the points (−2,0) and (1,0).Īnd don't forget the parabolas in the "legs down" orientation: Let's substitute x = 0 into the equation I just got to check if it's correct. Observe my graph passes through −3 on the y-axis. This is a quadratic function which passes through the x-axis at the required points. Now, we can write our function for the quadratic as follows (since if we solve the following for 0, we'll get our 2 intersection points): X = 1 (since the graph cuts the x-axis at x = 1.) X = −2 (since the graph cuts the x-axis at x = − 2) and We can see on the graph that the roots of the quadratic are: (We'll assume the axis of the given parabola is vertical.) Parabola cuts the graph in 2 places Sometimes it is easy to spot the points where the curve passes through, but often we need to estimate the points. Our job is to find the values of a, b and c after first observing the graph. We know that a quadratic equation will be in the form: The parabola can either be in "legs up" or "legs down" orientation. The graph of a quadratic function is a parabola. (Most "text book" math is the wrong way round - it gives you the function first and asks you to plug values into that function.)Ī quadratic function's graph is a parabola Often we have a set of data points from observations in an experiment, say, but we don't know the function that passes through our data points. This is a good question because it goes to the heart of a lot of "real" math.
#Quadratic equation graph how to#
I would like to know how to find the equation of a quadratic function from its graph, including when it does not cut the x-axis.
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