Name______Date______

Algebra 2 Notes Section 2-7

Curve Fitting with Quadratic Models

If the first differences are constant in a pattern of data, then the data could represent a linear function.

When the second differences are constant in a pattern of data, the data could represent a quadratic function.

x / 2 / 3 / 4 / 5 / 6
y / 6 / 12 / 20 / 30 / 42

Data Set 1

Find the first differences. This means the differences between successive y-values.

12–620–1230–2042–30

681012

Find the second differences. This means the differences between successive first differences.

8–610–812–10

222

Data Set 1 is a quadratic function.

x / 1 / 0 / 1 / 2 / 3
y / 0 / 1 / 2 / 9 / 28

Data Set 2

Find the first differences.

1–02–19–228–9

11719

Find the second differences.

1–17–119–7

0612

Data Set 2 is NOT a quadratic function.

Find the second differences. Determine whether each data set could represent a quadratic function.

x / 1 / 2 / 3 / 4 / 5
y / 1 / 3 / 7 / 13 / 21
x / 6 / 4 / 2 / 0 / 2
y / 8 / 5 / 2 / 5 / 8

First, we have to prepare the calculator by clearing stored entries and lists.

2nd MEM (above +) → 3: Clear Entries → Enter → Enter. “Done”

2nd MEM (above +) → 4: ClrAllLists → Enter → Enter. “Done”

Next, we have to turn on the diagnostic mode.

2nd CATALOG (above 0) scroll down to DiagnosticOn → Enter → Enter. “Done”

STAT → EDIT →1: Edit → ENTERSTAT → CALC →5: QuadReg → ENTER.

The closer the value of

R2 is to 1, the better the fit. fit.

Y= →\Y1: Input .397x2 -3.12x + 11.94

2nd GRAPH (TABLE); +: ∆Tbl = .5 (increment)

Y= →\Y1: Input 0.397x2 – 3.12x + 11.94

Set WINDOW: Xmin = 0, Xmax = 15, Xscl = 1

Ymin = 0, Ymax = 50, Yscl = 1 (due to rounding, the

2nd STAT PLOT (above Y =) → 1: Plot 1…On → ENTER. Y1 values will be

Highlight “Off” different)

GRAPH