Transcript Chapter 7

Chapter 7
Quadratic Equations
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A quadratic equation is one that can be written in
the form _____________________________
where a, b, c are real numbers and a ≠0.
The degree of a quadratic equation is ______.
E.g.
2 x  8  7 x
 x  3
1
3
2
2
 25
x  2  8x  x
2
2
2
The related function has equation ______________
It has a graph in the shape of a _______________.
Every quadratic equation has *two solutions (roots).
They may be:
a) ___________________________________
b) ___________________________________
c) ___________________________________
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Methods of Solving Quadratic Equations
With equation in form ax2 + bx + c =0
– Graphing: Graph related function y = ax2 + bx + c
and locate its real roots (x-intercepts)
On TI-83/84, use 2nd Calc 2: Zero
– Factoring: If possible, factor the expression.
Set each factor equal to zero and solve.
– Quadratic Formula:
x 
b 
b  4 ac
2
2a
4
Solve by graphing. Give answers to nearest tenth.
1)
2)
 x  1.8   0.2 x  1 
2
4( x  3)   x
2
2
5
Solve by factoring.
5 p  125  0
2
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Solve by factoring.
3m
2
 24m
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Solve by factoring.
1
6
 x
x
 
 11 7  77
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Solve the higher order equation by factoring.
x  13 x  36  0
4
2
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Solve the higher order equation by factoring.
x  4x  x  4  0
3
2
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Solve the rational equation by factoring.
x
2

1
x3
3
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Solve using the Quadratic Formula.
3x  5x  2  0
2
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Solve using the Quadratic Formula
(to two decimal places).
2 x  4 x  1
2
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For equations in the form ax2 + bx + c =0, the
discriminant is the value of ______________ .
(The expression under the radical in the Quadratic Formula.)
We can use the discriminant to determine the
character (number and type) of the roots of a
quadratic equation.
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Character of the Roots
• If b2 – 4ac > 0 and is a perfect square, the equation has
________________________________________.
• If b2 – 4ac > 0 and is NOT a perfect square, the equation
has _______________________________________.
• If b2 – 4ac = 0, the equation has
__________________________________________.
• If b2 – 4ac < 0, the equation has ________________
___________________________.
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Describe the number and type of roots.
1) x  x  1  0
2) 3 x  5 x  2  0
3) x  x  1  0
4) x  4 x  4  0
2
2
2
2
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Mathematical Modeling
In real world applications we often encounter numerical
data in the form of a table. The powerful mathematical
tool, regression analysis, can be used to analyze
numerical data. In general, regression analysis is a
process for finding a function that best fits a set of data
points.
In the next example, we use a linear model obtained by
using linear regression on a graphing calculator.
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Regression Notes
Regression: a process used to relate two quantitative
variables.
Independent variable: the x variable (or explanatory
variable)
Dependent variable: the y variable (or response variable)
To interpret the scatterplot, identify the following:
– Form
– Direction (for linear models)
– Strength
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Form
Form: the function that best describes the relationship
between the two variables.
Some possible forms would be linear, quadratic, cubic,
exponential, or logarithmic.
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Direction
Direction: a positive or negative direction can be found
when looking at linear regression lines only.
The direction is found by looking at the sign of the slope.
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Strength
Strength: how closely the points in the data are gathered
around the form.
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Making Predictions
Predictions should only be made for values of x within
the span of the x-values in the data set. Predictions
made outside the data set are called extrapolations,
which can be dangerous and ridiculous, thus
extrapolating is not recommended.
To make a prediction within the span of the x-values, hit 
then .
Next, arrow up or down  until the regression equation
appears in the upper-left hand corner then type in the xvalue and hit .
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Example of Linear Regression
Prices for emerald-shaped diamonds taken from an on-line
trader are given in the following table. Find the linear model
that best fits this data.
Weight (carats)
0.5
0.6
0.7
0.8
0.9
Price
$1,677
$2,353
$2,718
$3,218
$3,982
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Scatter Plots
Enter these values into the lists in a graphing calculator as
shown below .
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Scatter Plots
We can plot the data points in the previous example on a
Cartesian coordinate plane, either by hand or using a graphing
calculator. If we use the calculator, we obtain the following
plot:
Price of diamond
(thousands)
Weight (tenths of a carat)
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Example of Linear Regression
(continued)
Based on the scatterplot, the data appears to be linearly
correlated; thus, we can choose linear regression from the
statistics menu, we obtain the second screen, which gives the
equation of best fit.
The linear equation of best fit
is y = 5475x - 1042.9.
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Scatter Plots
We can plot the graph of our line of best fit on top of the
scatter plot:
y = 5475x - 1042.9
Price of emerald
(thousands)
Weight (tenths of a carat)
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Making a Prediction
Is it appropriate to use the model to predict the price of an
emerald-shaped diamond that weighs .75 carats? If so,
estimate the price.
Is it appropriate to use the model to predict the price of an
emerald-shaped diamond that weighs 2.7 carats? If so,
estimate the price.
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Quadratic Regression
A visual inspection of the plot of a data set might indicate that a
parabola would be a better model of the data than a straight line.
In that case, rather than using linear regression to fit a linear model
to the data, we would use quadratic regression on a graphing
calculator to find the function of the form y = ax2 + bx + c that best
fits the data.
From the  CALC menu, choose 5: QuadReg
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Example of Quadratic
Regression
An automobile tire manufacturer collected the data in the
table relating tire pressure x (in pounds per square inch) and
mileage (in thousands of miles.)
x
Mileage
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45
30
52
32
55
34
51
36
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Using quadratic regression
on a graphing calculator, find
the quadratic function that
best fits the data.
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Example of Quadratic Regression
(continued)
Enter the data in a graphing calculator and obtain the lists below.
Choose quadratic regression from the statistics menu and obtain
the coefficients as shown:
This means that the equation that best fits the data is:
y = -0.517857x2 + 33.292857x- 480.942857
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Example of Quadratic Regression
(continued)
Use the model to estimate the number of miles you could get
from tires inflated at a) 35 psi and b) 40 psi.
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Another Example of Modeling
The following table shows crop yields Y for various amounts of fertilizer
used, x, for 18 different equally sized plots. Assume that all values
are given to three significant digits.
Plot # 1
2
3
4
5
6
7
8
9
19 11 12 13 14 15 16 17 18
x
Fertilizer
(lbs/ ft2)
0
0
5
5
10 10 15 15 20 20 25 25 30 30 35 35 40 40
4
6
10 7
12 10 15 17 18 21 20 21 21 22 21 20 19 19
Y
Yield
(bushels)
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Example (continued)
1.
Use your calculator to graph a scatter plot of the data and
comment on the type of relationship that exists between the two
variables (the amount of fertilizer used , x, and the crop yield, y.)
It appears that the data follows a
quadratic relationship with a < 0.
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Example (continued)
2.
Use the calculator to find the quadratic function of best fit. Give
values to 4 significant digits. Sketch this function in the same
window as your scatter plot.
T he quadratic function of best fit to t his data i s
Y  x    0.01712 x  1.077 x  3.8 94
2
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Example (continued)
3.
Use the function to predict the optimal amount of fertilizer (in pounds)
to use and the crop yield (in bushels) when the optimal amount of
fertilizer is applied. Give values to 3 significant digits.
Use the graphing calculator and the graph of the quadratic model to
find the maximum point.
According to the model, if we apply 31.5 pounds of fertilizer
per 100 sq. feet, the crop yield will be 20.8 bushels.
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