1 Chapter 7 Quadratic Equations
Dec 17, 2015
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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.
22 8 7x x
23 25x
2 212 8
3x x x
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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 EquationsWith 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:
2 4
2
b b acx
a
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Solve by graphing. Give answers to nearest tenth.
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2
1) 1.8 0.2 1
2) 4( 3)
x x
x x
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Solve by factoring.
25 125 0p
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Solve by factoring.
23 24m m
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Solve by factoring.
1 6
11 7 77
xx
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Solve the higher order equation by factoring.
4 213 36 0x x
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Solve the higher order equation by factoring.
3 24 4 0x x x
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Solve using the Quadratic Formula.
23 5 2 0x x
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Solve using the Quadratic Formula (to two decimal places).
22 4 1x x
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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. 21) 1 0x x
23) 1 0x x
22) 3 5 2 0x x
24) 4 4 0x x
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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: 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
Regression Notes
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FormForm: 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 x-value 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) Price
0.5 $1,6770.6 $2,3530.7 $2,7180.8 $3,2180.9 $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
Price of diamond (thousands)
Weight (carats)
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:
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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:
Price of emerald (thousands)
Weight (carats)
y = 5475x - 1042.9
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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 0.75 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
28 45
30 52
32 55
34 51
36 47
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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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
YYield
(bushels)
4 6 10 7 12 10 15 17 18 21 20 21 21 22 21 20 19 19
Another Example of Modeling
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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.
Example (continued)
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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.
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The quadratic function of best fit to t
0.01712
his data i
1.077 3.8
s
94Y x x x
Example (continued)
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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.
Example (continued)
According to the model, if we apply 31.5 pounds of fertilizer
per 100 sq. feet, the crop yield will be 20.8 bushels.
Use the graphing calculator and the graph of the quadratic model to find the maximum point.