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Introduction Variables change, but constants remain the same. We need to understand how each term of an expression works in order to understand how changing.

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Presentation on theme: "Introduction Variables change, but constants remain the same. We need to understand how each term of an expression works in order to understand how changing."— Presentation transcript:

1 Introduction Variables change, but constants remain the same. We need to understand how each term of an expression works in order to understand how changing the value of a variable will affect the value of the expression. In this lesson, we will explore these relationships. 1 5.1.2: Interpreting Complicated Expressions

2 Key Concepts We can translate verbal expressions into algebraic expressions to analyze how changing values of coefficients, constants, or variables will affect the value of the expression. Changing the value of a variable will not change the value of a constant, just as changing the value of a constant will not change terms containing variables. It is important to always follow the correct order of operations. 2 5.1.2: Interpreting Complicated Expressions

3 Key Concepts, continued Simplifying expressions lets us more easily see how different terms interact with one another. Use the Distributive Property to multiply binomials. When presented with a polynomial in factored form, multiply the factors to see if the polynomial is a quadratic. Quadratic expressions are of the form ax 2 + bx + c, where a ≠ 0. 3 5.1.2: Interpreting Complicated Expressions

4 Key Concepts, continued Sometimes it is easier to work with the factored form of an expression. For instance, since we know that a positive number multiplied by a positive number is always a positive number, it is easier to determine values of a variable that make an expression positive by finding the values that make all of its factors positive. We can use similar logic to determine the values that make an expression negative or equal to 0. 4 5.1.2: Interpreting Complicated Expressions

5 Common Errors/Misconceptions forgetting to distribute when multiplying binomials incorrectly following the order of operations incorrectly translating verbal expressions 5 5.1.2: Interpreting Complicated Expressions

6 Guided Practice Example 3 The length of each side of a square is increased by 2 centimeters. How does the perimeter change? How does the area change? 6 5.1.2: Interpreting Complicated Expressions

7 Guided Practice: Example 3, continued 1.Find an expression for the perimeter of the original square. Let x be the length of one side of the square in centimeters. Multiply each side length by 4 since a square has four sides of equal length. The perimeter of the square is 4x centimeters. 7 5.1.2: Interpreting Complicated Expressions

8 Guided Practice: Example 3, continued 2.Find an expression for the perimeter of the new square and compare the new perimeter to the original perimeter. One side of the new square is x + 2 centimeters. The new perimeter is 4(x + 2) or 4x + 8 centimeters. Next, subtract the old perimeter from the new perimeter to determine the difference. 4x + 8 – 4x = 8 The new perimeter is 8 centimeters longer than the original perimeter. 8 5.1.2: Interpreting Complicated Expressions

9 Guided Practice: Example 3, continued 3.Find an expression for the area of the original square. The area of the original square is x 2 square centimeters. 9 5.1.2: Interpreting Complicated Expressions

10 Guided Practice: Example 3, continued 4.Find an expression for the area of the new square and compare the new area to the original area. The new area is (x + 2) 2 or x 2 + 4x + 4 square centimeters. Subtract the old area from the new area to determine the difference. x 2 + 4x + 4 – x 2 = 4x + 4 The new area is 4x + 4 square centimeters larger than the old area. 10 5.1.2: Interpreting Complicated Expressions ✔

11 Guided Practice: Example 3, continued 11 5.1.2: Interpreting Complicated Expressions

12 Guided Practice Example 4 A car’s total stopping distance in feet depends on many factors, but can be approximated by the expression, for which x is the speed of the car in miles per hour. Is this expression quadratic? What effect does doubling the car’s speed from 10 mph to 20 mph have on the total stopping distance? 12 5.1.2: Interpreting Complicated Expressions

13 Guided Practice: Example 4, continued 1.Determine whether the expression is quadratic. This expression is quadratic because it can be written in the form ax 2 + bx + c, with and c = 0. 13 5.1.2: Interpreting Complicated Expressions

14 Guided Practice: Example 4, continued 2.Determine the stopping distance of a car traveling 10 mph. Substitute 10 for x to find the total stopping distance of a car traveling 10 mph. The stopping distance of a car traveling 10 mph is approximately 16.26 feet. 14 5.1.2: Interpreting Complicated Expressions

15 Guided Practice: Example 4, continued 3.Determine the stopping distance of a car traveling 20 mph. Substitute 20 for x to find the total stopping distance of a car traveling 20 mph. The stopping distance of a car traveling 20 mph is approximately 43.05 feet. 15 5.1.2: Interpreting Complicated Expressions

16 Guided Practice: Example 4, continued 4.Compare the stopping distances for a car traveling 10 mph and a car traveling 20 mph. When the speed of the car doubles from 10 mph to 20 mph, the stopping distance almost triples from about 16 feet to about 43 feet. 16 5.1.2: Interpreting Complicated Expressions ✔

17 Guided Practice: Example 4, continued 17 5.1.2: Interpreting Complicated Expressions

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