Quadratic Models without QuadReg Julie Graves NCSSM and Philips Exeter Academy June 2014

A quadratic data set: x y

The curvature of the scatterplot suggests a downward opening parabola. For any such parabola, the y values are increasing to the left of the vertex and are decreasing to the right of the vertex. Even though the scatterplot does not allow us to guess the location of the vertex with any precision, we can study the changes in y values to find the x coordinate of the vertex.

x y Mid x

We want to use this information to predict the x value at which would change sign.

The equation of the least squares regression line is From the equation, we can conclude that the sign of changes when the x-value is 10.75/2.5, or x=4.3. Thus, we have reason to believe that a parabolic model has its vertex located at x=4.3; the equation of such a model has the form

The quadratic equation implies that there is a linear relationship between and y; we can find appropriate values for each of the constants a and k using linear regression on ordered pairs Linear regression yields a slope of -2.5 and an intercept equal to This means that we have found a quadratic model for our data; it is

We can use a similar strategy to find an exponential model for some data. x y

For an exponential model, the y coordinate of the horizontal asymptote represents the y value at which y values stop changing. We want to use the data to predict the y value at which would be zero.

x y Midy

The equation of the least squares regression line is From this equation, we can conclude that would be zero when the y-value is 6.613/0.0968, or Thus, we have evidence from the data itself that the horizontal asymptote is located at about

This means that an exponential model for the data has the form To find appropriate values for the constants a and b, we can either  fit an linear model to ordered pairs of the form (x, ln(y )), which yields or  fit an exponential model to ordered pairs (x, y ) which yields