Transcript Chapter 7 Notes

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
ax  bx  c  0
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– Graphing: Graph related function y  ax 2  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
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Solve by graphing on the calculator. Give answers to
nearest tenth.
1)
2)
 x  1.8   0.2 x  1 
2
4( x  3)   x
2
2
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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 using the Quadratic Formula.
3x  5x  2  0
2
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Solve using the Quadratic Formula (Give answers to two decimal
places.)
2 x  4 x  1
2
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For equations in the form ax  bx  c  0 , the discriminant
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is the value of ______________ .
(This is 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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First find the value of the discriminant; then use it to
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.
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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 (carats)
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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 scatterplot:
y = 5475x - 1042.9
Price of emerald
(thousands)
Weight (carats)
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Making a Prediction
Is it appropriate to use the model to predict the price of an emeraldshaped diamond that weighs 0.75 carats? If so, estimate the price.
Is it appropriate to use the model to predict the price of an emeraldshaped 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.
Round values to 6 decimal places.
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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 STAT Calc 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 (in bushels), for various amounts
of fertilizer used, x (in lbs/100 ft2), for 18 different equally sized plots.
Plot # 1
2
3
4
5
6
7
8
9
19 11 12 13 14 15 16 17 18
x
Fertilizer
(lbs/
100ft2)
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
per 100ft2) 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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