5–Minute Check 3 Evaluate |x – 2y| – |2x – y| – xy if x = –2 and y = 7. A.–9 B.9 C.19 D.41

5–Minute Check 3 Evaluate |x – 2y| – |2x – y| – xy if x = –2 and y = 7. A.–9 B.9 C.19 D.41

5–Minute Check 4 Factor 8xy 2 – 4xy. A.2x(4xy 2 – y) B.4xy(2y – 1) C.4xy(y 2 – 1) D.4y 2 (2x – 1)

5–Minute Check 4 Factor 8xy 2 – 4xy. A.2x(4xy 2 – y) B.4xy(2y – 1) C.4xy(y 2 – 1) D.4y 2 (2x – 1)

5–Minute Check 5 A. B. C. D.

5–Minute Check 5 A. B. C. D.

Vocabulary interval notation function function notation independent variable dependent variable piecewise-defined function relevant domain

Key Concept 1

Example 2 Use Interval Notation A. Write –2 ≤ x ≤ 12 using interval notation. The set includes all real numbers greater than or equal to –2 and less than or equal to 12. Answer:

Example 2 Use Interval Notation A. Write –2 ≤ x ≤ 12 using interval notation. The set includes all real numbers greater than or equal to –2 and less than or equal to 12. Answer: [–2, 12]

Example 2 Use Interval Notation B. Write x > –4 using interval notation. The set includes all real numbers greater than –4. Answer:

Example 2 Use Interval Notation B. Write x > –4 using interval notation. The set includes all real numbers greater than –4. Answer: (–4, )

Example 2 Use Interval Notation C. Write x < 3 or x ≥ 54 using interval notation. The set includes all real numbers less than 3 and all real numbers greater than or equal to 54. Answer:

Example 2 Use Interval Notation C. Write x < 3 or x ≥ 54 using interval notation. The set includes all real numbers less than 3 and all real numbers greater than or equal to 54. Answer:

Example 2 Write x > 5 or x < –1 using interval notation. A. B. C.(–1, 5) D.

Example 2 Write x > 5 or x < –1 using interval notation. A. B. C.(–1, 5) D.

Key Concept 3

Key Concept 3a

Example 3 Identify Relations that are Functions B. Determine whether the table represents y as a function of x. Answer:

Example 3 Identify Relations that are Functions B. Determine whether the table represents y as a function of x. Answer: No; there is more than one y-value for an x-value.

Example 3 Identify Relations that are Functions C. Determine whether the graph represents y as a function of x. Answer:

Example 3 Identify Relations that are Functions C. Determine whether the graph represents y as a function of x. Answer: Yes; there is exactly one y-value for each x- value. Any vertical line will intersect the graph at only one point. Therefore, the graph represents y as a function of x.

Example 3 Identify Relations that are Functions D. Determine whether x = 3y 2 represents y as a function of x. To determine whether this equation represents y as a function of x, solve the equation for y. x= 3y 2 Original equation Divide each side by 3. Take the square root of each side.

Example 3 Identify Relations that are Functions Answer: This equation does not represent y as a function of x because there will be two corresponding y-values, one positive and one negative, for any x-value greater than 0. Let x = 12.

Example 3 Identify Relations that are Functions Answer: No; there is more than one y-value for an x-value. This equation does not represent y as a function of x because there will be two corresponding y-values, one positive and one negative, for any x-value greater than 0. Let x = 12.

Example 3 Determine whether 12x 2 + 4y = 8 represents y as a function of x. A.Yes; there is exactly one y-value for each x-value. B.No; there is more than one y-value for an x-value.

Example 3 Determine whether 12x 2 + 4y = 8 represents y as a function of x. A.Yes; there is exactly one y-value for each x-value. B.No; there is more than one y-value for an x-value.

Example 4 Find Function Values A. If f (x) = x 2 – 2x – 8, find f (3). To find f (3), replace x with 3 in f (x) = x 2 – 2x – 8. f (x)= x 2 – 2x – 8Original function f (3)= 3 2 – 2(3) – 8 Substitute 3 for x. = 9 – 6 – 8 Simplify. = –5 Subtract. Answer:

Example 4 Find Function Values A. If f (x) = x 2 – 2x – 8, find f (3). To find f (3), replace x with 3 in f (x) = x 2 – 2x – 8. f (x)= x 2 – 2x – 8Original function f (3)= 3 2 – 2(3) – 8 Substitute 3 for x. = 9 – 6 – 8 Simplify. = –5 Subtract. Answer:–5

Example 4 Find Function Values B. If f (x) = x 2 – 2x – 8, find f (–3d). To find f (–3d), replace x with –3d in f (x) = x 2 – 2x – 8. f (x)= x 2 – 2x – 8Original function f (–3d)= (–3d) 2 – 2(–3d) – 8Substitute –3d for x. = 9d 2 + 6d – 8 Simplify. Answer:

Example 4 Find Function Values B. If f (x) = x 2 – 2x – 8, find f (–3d). To find f (–3d), replace x with –3d in f (x) = x 2 – 2x – 8. f (x)= x 2 – 2x – 8Original function f (–3d)= (–3d) 2 – 2(–3d) – 8Substitute –3d for x. = 9d 2 + 6d – 8 Simplify. Answer:9d 2 + 6d – 8

Example 4 Find Function Values C. If f (x) = x 2 – 2x – 8, find f (2a – 1). To find f (2a – 1), replace x with 2a – 1 in f (x) = x 2 – 2x – 8. f (x)= x 2 – 2x – 8Original function f (2a – 1)= (2a – 1) 2 – 2(2a – 1) – 8 Substitute 2a – 1 for x. = 4a 2 – 4a + 1 – 4a + 2 – 8 Expand (2a – 1) 2 and 2(2a – 1). = 4a 2 – 8a – 5Simplify. Answer:

Example 4 Find Function Values C. If f (x) = x 2 – 2x – 8, find f (2a – 1). To find f (2a – 1), replace x with 2a – 1 in f (x) = x 2 – 2x – 8. f (x)= x 2 – 2x – 8Original function f (2a – 1)= (2a – 1) 2 – 2(2a – 1) – 8 Substitute 2a – 1 for x. = 4a 2 – 4a + 1 – 4a + 2 – 8 Expand (2a – 1) 2 and 2(2a – 1). = 4a 2 – 8a – 5Simplify. Answer:4a 2 – 8a – 5

Example 4 If, find f (6). A. B. C. D.

Example 4 If, find f (6). A. B. C. D.

Example 5 Find Domains Algebraically A. State the domain of the function. Because the square root of a negative number cannot be real, 4x – 1 ≥ 0. Therefore, the domain of g(x) is all real numbers x such that x ≥, or. Answer:

Example 5 Find Domains Algebraically A. State the domain of the function. Because the square root of a negative number cannot be real, 4x – 1 ≥ 0. Therefore, the domain of g(x) is all real numbers x such that x ≥, or. Answer:all real numbers x such that x ≥, or

Example 5 Find Domains Algebraically B. State the domain of the function. When the denominator of is zero, the expression is undefined. Solving t 2 – 1 = 0, the excluded values in the domain of this function are t = 1 and t = –1. The domain of this function is all real numbers except t = 1 and t = –1, or. Answer:

Example 5 Find Domains Algebraically B. State the domain of the function. When the denominator of is zero, the expression is undefined. Solving t 2 – 1 = 0, the excluded values in the domain of this function are t = 1 and t = –1. The domain of this function is all real numbers except t = 1 and t = –1, or. Answer:

Example 5 Find Domains Algebraically C. State the domain of the function. Answer: This function is defined only when 2x – 3 > 0. Therefore, the domain of f (x) is or.

Example 5 Find Domains Algebraically C. State the domain of the function. Answer: or This function is defined only when 2x – 3 > 0. Therefore, the domain of f (x) is or.

Example 5 State the domain of g (x) =. A. or [4, ∞ ) B. or [–4, 4] C. or (−, −4] D.

Example 5 State the domain of g (x) =. A. or [4, ∞ ) B. or [–4, 4] C. or (−, −4] D.

Example 6 A. FINANCE Realtors in a metropolitan area studied the average home price per square foot as a function of total square footage. Their evaluation yielded the following piecewise- defined function. Find the average price per square foot for a home with the square footage of 1400 square feet. Evaluate a Piecewise-Defined Function

Example 6 Evaluate a Piecewise-Defined Function Because 1400 is between 1000 and 2600, use to find p(1400). Subtract. = 85Simplify. Function for 1000 ≤ a < 2600 Substitute 1400 for a.

Example 6 Answer: Evaluate a Piecewise-Defined Function According to this model, the average price per square foot for a home with a square footage of 1400 square feet is $85.

Example 6 Answer: $85 per square foot Evaluate a Piecewise-Defined Function According to this model, the average price per square foot for a home with a square footage of 1400 square feet is $85.

Example 6 B. FINANCE Realtors in a metropolitan area studied the average home price per square foot as a function of total square footage. Their evaluation yielded the following piecewise-defined function. Find the average price per square foot for a home with the square footage of 3200 square feet. Evaluate a Piecewise-Defined Function

Example 6 Evaluate a Piecewise-Defined Function Because 3200 is between 2600 and 4000, use to find p(3200). Function for 2600 ≤ a < Simplify. Substitute 3200 for a.

Example 6 Answer: Evaluate a Piecewise-Defined Function According to this model, the average price per square foot for a home with a square footage of 3200 square feet is $104.

Example 6 Answer: $104 per square foot Evaluate a Piecewise-Defined Function According to this model, the average price per square foot for a home with a square footage of 3200 square feet is $104.

Example 6 ENERGY The cost of residential electricity use can be represented by the following piecewise function, where k is the number of kilowatts. Find the cost of electricity for 950 kilowatts. A.$47.50 B.$48.00 C.$57.50 D.$76.50

Example 6 ENERGY The cost of residential electricity use can be represented by the following piecewise function, where k is the number of kilowatts. Find the cost of electricity for 950 kilowatts. A.$47.50 B.$48.00 C.$57.50 D.$76.50