Finding the
Equation of a
Quadratic
Dm
Timesec
Using QUADDATA
recorded by rolling a ball
up and down a ramp. The
distance is measured from
an arbitrary starting point.
Distance from the starting
point toward the motion
detector is positive.
© 2002 Jerel L. Welker, Lincoln High School
Permission is granted to copy and distribute
with copyright notation attached.
Finding a Quadratic Equation in Vertex Form y=a(x-h)2 + k
Step 1: Find the Vertex
Use the TRACE function to
find the vertex of the parabola.
The vertex occurs at (h, k). The parabola has a vertex at
(1.65 sec, 0.463 m).
The equation is y = a(x - 1.65 sec)2 + 0.463 m
Finding a Quadratic Equation in Vertex Form y=a(x-h)2 + k
Step 2: Find the value of ‘a’
Use the TRACE function to
find another point on the
parabola.
Substitute for (x, y) in the equation y = a(x - 1.65 sec)2 +
0.463 m. Solve algebraically for ‘a’.
-0.583 m = a(2.73 sec - 1.65 sec)2 + 0.463 m
-0.583 m - 0.463 m = a(2.73 sec - 1.65 sec)2
(-0.583 m - 0.463 m)
=a
2
(2.73 sec - 1.65 sec)
a = -0.897 m/sec2
Finding a Quadratic Equation in Vertex Form y=a(x-h)2 + k
Step 3: Graph the Equation
Substituting for ‘a’ in the equation y = a(x - 1.65 sec)2 +
0.463 m, one gets y = -0.897 m/sec2 (x - 1.65 sec)2 + 0.463 m.
Use the function grapher to graph:
y = -0.897 m/sec2 (x - 1.65 sec)2 + 0.463 m.
If you want to try and improve the
fit, use different values of a, h,
and k to see how each changes the
graph.
Finding a Quadratic Equation in Vertex Form y=a(x-h)2 + k
Step 4: Write the mathematical model.
The mathematical model is:
D m = -0.90 m/sec2 (T sec - 1.65 sec)2 + 0.46 m.
Step 5: Explain the mathematical model.
The ball reached its point furthest from the starting
point (origin) or closest to the motion detector 1.65
sec after data collection started. This point was 0.46
meters from the starting point. One-half of the
acceleration was -0.90 m/sec2. Up the ramp was
positive, so the acceleration was down the ramp.
Finding a Quadratic Equation in Factored Form y=a(x-r)(x-s)
Before we begin Factored Form:
Turn off the Vertex equation in the function grapher by
placing the cursor on the = sign.
Press ENTER to remove
the highlight on the = sign.
The function is stored in
memory but is not graphed
unless turned on.
Finding a Quadratic Equation in Factored Form y=a(x-r)(x-s)
Step 1: Find the x-intercepts, roots, or zeroes of the polynomial
using the TRACE function.
The values of ‘r’ and ‘s’ are the x-intercepts or where the
graph crosses the x-axis. Find the closest point, average, or
estimate the roots.
Note that when r = 0.9
y is very close to 0.
r = 0.9 sec
The value of ‘s’ is between 2.34 and
2.37 but closer to 2.37.
Estimate s = 2.36 sec
Finding a Quadratic Equation in Factored Form y=a(x-r)(x-s)
Step 2: Find the ‘a’ value.
Substitute values of ‘r’ and ‘s’ into the equation. Choose a
third point for ‘x’ and ‘y’ and solve for ‘a’. The vertex is a
good choice for the third point.
y = a (x - 0.9 sec)(x - 2.36 sec)
Substituting the vertex (1.65, 0.463):
0.463m= a(1.65sec - 0.9sec)(1.65sec- 2.36sec)
Solving for a:
a=
0.463m
((1.65sec - 0.9sec)(1.65sec- 2.36sec))
a = -0.869 m/sec2
Finding a Quadratic Equation in Factored Form y=a(x-r)(x-s)
Step 3: Graph the equation.
y = -0.869 m/sec2 (x - 0.9 sec)(x - 2.36 sec)
Place the equation in Y2 as
Y1 has been turned off.
If you want to try and improve the
fit, use different values of ‘a’, ‘r’,
and ‘s’ to see how each changes
the graph.
Finding a Quadratic Equation in Factored Form y=a(x-r)(x-s)
Step 4: Write a Mathematical Model.
D m = -0.87 m/sec2 (T sec - 0.9 sec)(T sec - 2.36 sec)
Step 5: Explain the Mathematical Model.
The ball passes the point marked as the origin 0.9
seconds after data collection started. The ball was
moving in a positive direction (up the ramp). 2.36
seconds after data collection started, the ball passed the
origin on its way down the ramp. One-half of the
acceleration is - 0.87 m/sec2 . Acceleration is in a negative
direction (down the ramp).
Finding a Quadratic Equation in Quadratic Form y=ax2 + bx + c
Before we begin finding Quadratic Form!
Turn off the Vertex equation in the function grapher by
placing the cursor on the = sign.
Press ENTER to remove
the highlight on the = sign.
The function is stored in
memory but is not graphed
unless turned on.
Finding a Quadratic Equation in Quadratic Form y=ax2 + bx + c
Quadratic form allows for practice expanding the factored and
vertex forms of a parabola.
Step 1: Expand (multiply out) the equation in factored form.
D m = -0.87 m/sec2 (T sec - 0.9 sec)(T sec - 2.36 sec)
D m = -0.87 (T2 - 2.36T - 0.9T + 2.124)
D m = -0.87 (T2 - 3.26T + 2.124)
D m = -0.87 m/sec2 T2 + 2.84 m/secT - 1.85 m
Finding a Quadratic Equation in Quadratic Form y=ax2 + bx + c
Step 2: Graph the model.
D m = -0.87 m/sec2 T2 + 2.84 m/secT - 1.85 m
Step 3: Explain the model.
One-half of the acceleration is -0.87 m/sec2 (down the ramp).
If the ball had been in continuous motion, it would have had
a velocity of 2.84 m/sec when data collection began (t=0).
The ball would have been 1.85 meters away from the
starting point in a negative direction when data collection
started.
Finding a Quadratic Equation in Quadratic Form y=ax2 + bx + c
Finding a second quadratic form from vertex form of the parabola.
Step 1: Expand the vertex form of the parabola.
D m = -0.90 m/sec2 (T sec - 1.65 sec)2 + 0.46 m.
D m = -0.90 (T 2 - 2*1.65*T + 1.65 2) + 0.46
D m = -0.90 (T 2 - 3.30T + 2.7225) + 0.46
D m = -0.90T 2 + 2.97T - 2.45025 + 0.46
D m = -0.90 m/sec2 T 2 + 2.97 m/secT - 1.99 m
Finding a Quadratic Equation in Quadratic Form y=ax2 + bx + c
Step 2: Graph the expanded vertex form of the parabola.
D m = -0.90 m/sec2 T 2 + 2.97 m/secT - 1.99 m
Step 3: Explain the mathematical model.
One-half of the accleration is -.90 m/sec2 (down the ramp). If
the ball had been in continuous motion, the initial velocity
would have been 2.97m/sec when data collection began (t=0).
The position of the ball at t=0 would have been -1.99m (below
the origin on the ramp).
Using the Quadratic Regression
y=ax2 + bx + c
Remember to turn off the equation in Y4!!
Students who skip immediately to the quadratic regression will find
that it does not fit well as a result of the erroneous data collected
before the ball started moving and after the ball was stopped.
The erroneous regression is shown graphed below.
Using the Quadratic Regression
y=ax2 + bx + c
Step 1: Choose an appropriate part of the graph for a regression.
From the PRGM menu, select the
CHOOSE program.
Press ENTER to run the program from
the HOME screen.
A descriptive message will appear
asking you to use the cursor keys to
select the left and right sides of the
data. The program will keep the data
between the two selected points.
Using the Quadratic Regression
y=ax2 + bx + c
Step 1: Choose an appropriate part (con’t).
Use the arrow keys to move the cursor
along the data to a point on the left side
of the data to be saved. All points to the
left of this point will not be used in the
regression. Press ENTER to continue.
Continue moving the cursor to a point
on the right side of the data to be
saved. All data between these two
points will be used in the regression.
Points to the right of this point will not
be used. Press ENTER to continue.
Using the Quadratic Regression
y=ax2 + bx + c
Step 1: Choose an appropriate part (con’t).
The original data set will remain in
L1 and L2. The selected points will
be copied to L3 and L4 as shown in
the message box.
Press ENTER to see the selected
data set graphed on the screen in a
new window.
Using the Quadratic Regression
y=ax2 + bx + c
Step 2: Find a Quadratic Regression.
Choose the QuadReg regression
from the STAT - CALC menu
The data for this regression is in L3
and L4. Enter these lists with
comma’s.
From the VARS - Y-Vars
Function menu, choose Y5 from
the list. The regression will be
stored in Y5. Press ENTER to
find the regression.
Using the Quadratic Regression
y=ax2 + bx + c
Step 2: Find a Quadratic Regression (con’t).
The regression in quadratic form
will be displayed. If DiagnosticOn
is selected, the correlation
coefficient will also be displayed.
Step 3: Graph the Quadratic Regression.
Press GRAPH to display the
graph with the selected data in
L3 and L4.
Using the Quadratic Regression
y=ax2 + bx + c
Step 4: Graph the regression with the original data set.
Choose STAT PLOT (2nd Y=).
Turn off Stat Plot 2 and turn Stat
Plot 1 back on. Note that Stat Plot
one graphs lists L1 and L2. Stat Plot
2 was turned on by the CHOOSE
program and graphs L3 and L4.
Set the WINDOW to show the
original data set or press
ZoomStat (Zoom-9) to view the
original window.
y=ax2 + bx + c
Using the Quadratic Regression
Step 4: Graph the regression (con’t).
Press GRAPH to display the graph
in the selected window.
The equation graphed is shown in
Y5 of the function grapher.
Using the Quadratic Regression
y=ax2 + bx + c
Step 5: Write the mathematical model.
D m = -0.87 m/sec2 T 2 + 2.83 m/secT - 1.84 m
Step 6: Explain the mathematical model.
One-half of the accleration is -.87m/sec2 (down the ramp). If the
ball had been in continuous motion, the initial velocity would
have been 2.83m/sec when data collection began (t=0). The
position of the ball at t=0 would have been -1.84m (below the
origin on the ramp).
The End