## I1.07: Section 5 Part 1

### Section 5: Using Models.xls to find a quadratic model

If we wish to find a model for the data on the right, this quick scatter plot of the data shows that a straight line will not be sufficient. Such parabolic shapes (similar to the path of a thrown ball) is instead represented mathematically by a quadratic formula, in which the term that contains the input variable x is squared.

There are several ways to write a quadratic formula, all of which can make the same curves. For use in fitting data, the best kind of quadratic formula is y = a (x h) 2 + v, where the parameters h and v are the x and y coordinates of the vertex of the parabola (that is, its highest or lowest point), and a is a “shape” parameter that determines how sharply (and in which direction) the parabola bends.

The quadratic formula pattern that is most convenient for fitting models: $y=a\cdot{{(x-h)}^{2}}+v$

a is a “shape” parameter controlling how much (and in which direction) the parabola bends

h is the x coordinate of the vertex (its horizontal distance from the origin)

v is the y coordinate of the vertex (its vertical distance from the origin)

Examples: $y=3\cdot{{(x-2)}^{2}}+8$   $y=2.5\cdot{{(x-7)}^{2}}-33.4$   $y=-2\cdot{{(x+3.5)}^{2}}+37$

 time height x y 0 0.0 1 8.6 2 16.8 3 24.4 4 31.5 5 37.4 6 43.8 7 48.9 8 54.4 9 58.9 10 63.3 11 66.5 12 69.7 13 72.2 14 74.5 15 76.2 16 77.8 17 78.1 18 79.0 19 78.5 20 78.0 21 76.9 22 75.6 23 73.0 24 70.3 25 67.3 26 63.7 27 60.0 28 55.0
Examples of graphs of various quadratic formulas
$y=3\cdot{{(x-2)}^{2}}+8$ $y=-1.2\cdot{{(x-2.5)}^{2}}-50$ $y=18\cdot{{(x+3)}^{2}}-158$ $y=3\cdot{{(x+7)}^{2}}-300$

Example 5: For each of the formulas above, state the location of the vertex of the parabola formed.

Solution: Since the vertex is at (h,v) when a formula is expressed in the form,      the coordinates for the vertices are: (2, 8) (2.5, −50) (−3, −158) (−7, −300)

Note that the sign of the x vertex coordinate is the opposite of the sign that the same number has in the formula, since the h value is subtracted when forming the formula.

Quadratic models are somewhat more complicated than linear ones, as is indicated by the fact that a quadratic model has three parameters instead of two. But there is really very little difference in the fitting process from what is done for straight lines: [1] put the data in the appropriate worksheet, [2] spread the C3:E3 formulas down beside the data, [3] make a graph and adjust the vertex (instead of the intercept) and the shape (instead of the slope) until the model and the data match, and [5] write down the formula or use it to predict any values you have been asked for.