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Quadratic Applications
Section 2.1A Precalculus PreAP/Dual, Revised Β©2017 1/17/ :32 AM Β§2.1A: Quadratic Applications
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Steps for Completing the Square
Put terms with variables on one side and CONSTANT to the other side Make sure the side is in DESCENDING order Standard Form, π¨ π π +π©π+πͺ=π where π¨=π Identify the coefficient which is raised to the first power, DIVIDE the term by 2, and SQUARE the number (it will always be positive) ADD to both sides to the equation Put the equation into FACTORED form (Vertex Form) SQUARE ROOT both sides and cancel the binomial Solve for π and check 1/17/ :32 AM Β§2.1A: Quadratic Applications
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Β§2.1A: Quadratic Applications
Example 1 Given π= π π +ππβπ, convert to Vertex Form 1/17/ :32 AM Β§2.1A: Quadratic Applications
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Β§2.1A: Quadratic Applications
Example 2 Given π=π π π βππ+π, convert to Vertex Form 1/17/ :32 AM Β§2.1A: Quadratic Applications
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Β§2.1A: Quadratic Applications
Example 2 Given π=π π π βππ+π, convert to Vertex Form 1/17/ :32 AM Β§2.1A: Quadratic Applications
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Β§2.1A: Quadratic Applications
Your Turn Given π π βπππ=βπ, convert to Vertex Form 1/17/ :32 AM Β§2.1A: Quadratic Applications
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Using Technology to Solve
When a functionβs graph intersects the π-axis and the π-value is a zero, it is called a zero, solution, or root. In the real-world, it is also called the π-intercept. The maximum or minimum will establish the vertex When units are in feet, the equation is π=βππ π π +ππ+π where π is in time and π is initial height When units are in meters, the equation is π=βπ.π π π +ππ+π where π is in time and π is initial height 1/17/ :32 AM Β§2.1A: Quadratic Applications
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Where does βπππ π and βπ.ππ π Come From?
Gravity pulls objects toward the center of the earth (βdownβ to us) at an acceleration of 32 feet per sec2 (American measure) or 9.8 meters per sec2 (metric measure). AVERAGE acceleration per second is what is used. An objectβs velocity will be greater at the end of the one-second interval than at the beginning of the interval. If acceleration at π=π is 0 and at π=π is βπ.π, then the average is βπ.π in that interval. 1/17/ :32 AM Β§2.1A: Quadratic Applications
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Β§2.1A: Quadratic Applications
Example 3 Using the graphing calculator, identify the zeros, π-intercept, and vertex for the equation, π=π π π βππβπ. 1/17/ :32 AM Β§2.1A: Quadratic Applications
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Β§2.1A: Quadratic Applications
Example 4 Suppose a ball is thrown upwards from a height of 5 feet with an initial velocity of ππ ππ πππ . Write an equation relating to the time (π) in seconds and height (π) of the ball in feet. Find the height of the ball after 1.5 seconds. When does the ball reach its maximum height? What is the maximum height? When will the ball be 17 feet in the air? How long will it take for the ball to hit the ground? 1/17/ :32 AM Β§2.1A: Quadratic Applications
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Β§2.1A: Quadratic Applications
Example 4a Suppose a ball is thrown upwards from a height of 5 feet with an initial velocity of ππ ππ πππ . Write an equation relating to the time (π) in seconds and height (π) of the ball in feet. 1/17/ :32 AM Β§2.1A: Quadratic Applications
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Β§2.1A: Quadratic Applications
Example 4b Suppose a ball is thrown upwards from a height of 5 feet with an initial velocity of ππ ππ πππ . b) Find the height of the ball after 1.5 seconds. 1/17/ :32 AM Β§2.1A: Quadratic Applications
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Β§2.1A: Quadratic Applications
Example 4c Suppose a ball is thrown upwards from a height of 5 feet with an initial velocity of ππ ππ πππ . c) When does the ball reach its maximum height? 1/17/ :32 AM Β§2.1A: Quadratic Applications
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Β§2.1A: Quadratic Applications
Example 4d Suppose a ball is thrown upwards from a height of 5 feet with an initial velocity of ππ ππ πππ . d) What is the maximum height? 1/17/ :32 AM Β§2.1A: Quadratic Applications
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Β§2.1A: Quadratic Applications
Example 4e Suppose a ball is thrown upwards from a height of 5 feet with an initial velocity of ππ ππ πππ . e) When will the ball be 17 feet in the air? 1/17/ :32 AM Β§2.1A: Quadratic Applications
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Β§2.1A: Quadratic Applications
Example 4f Suppose a ball is thrown upwards from a height of 5 feet with an initial velocity of ππ ππ πππ . f) How long will it take for the ball to hit the ground? 1/17/ :32 AM Β§2.1A: Quadratic Applications
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Β§2.1A: Quadratic Applications
Example 5 Johanna threw a water balloon upward at a speed of 10 m/sec while standing on the roof of a building that is 12 meters high. Write an equation relating the time (π) in seconds and height (π) of the balloon in meters. What was the height of the balloon after 2 seconds? When will the ball reach its maximum height? What is the maximum height? When will the ball be 5 meters in the air? Assume the balloon did not land on the roof. How long it took the balloon to reach the ground? 1/17/ :32 AM Β§2.1A: Quadratic Applications
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Β§2.1A: Quadratic Applications
Your Turn Jamie threw a stone upward at 10 meters/sec while standing on a cliff 40 meters above the ground. Write an equation relating the time (π) in seconds and height (π) of the balloon in meters. What was the height of the balloon after 1.25 seconds? When will the ball reach its maximum height? What is the maximum height? When will the ball be 10 meters in the air? How long will it take for the stone to reach the water at the bottom of the cliff? 1/17/ :32 AM Β§2.1A: Quadratic Applications
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Β§2.1A: Quadratic Applications
Example 6 The path of a baseball after being hit is given by the function π π =βπ.ππππ π π +π+π where π π is the height of the baseball (in feet) and π is the horizontal distance from home plate (in feet). What is the maximum height of the baseball? (calc) 1/17/ :32 AM Β§2.1A: Quadratic Applications
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Β§2.1A: Quadratic Applications
Example 6 The path of a baseball after being hit is given by the function π π =βπ.ππππ π π +π+π where π π is the height of the baseball (in feet) and π is the horizontal distance from home plate (in feet). What is the maximum height of the baseball? (calc) 1/17/ :32 AM Β§2.1A: Quadratic Applications
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Β§2.1A: Quadratic Applications
Example 6 The path of a baseball after being hit is given by the function π π =βπ.ππππ π π +π+π where π π is the height of the baseball (in feet) and π is the horizontal distance from home plate (in feet). What is the maximum height of the baseball? (calc) 1/17/ :32 AM Β§2.1A: Quadratic Applications
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Β§2.1A: Quadratic Applications
Your Turn The path of a punted football is given by π π =β ππ ππππ π π + π π π+π.π where π π in the height in feet and π is the horizontal distance (in feet) from the point at which the ball is punted. What is the maximum height of the punt? Round to 4 decimal places. 1/17/ :32 AM Β§2.1A: Quadratic Applications
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Β§2.1A: Quadratic Applications
Assignment Worksheet 1/17/ :32 AM Β§2.1A: Quadratic Applications
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