1. Introduction A function is a relation in which every element of the domain is paired with exactly one element of the range; that is, for every value of x, there is exactly one value of y. When writing functions based on real-world information, first we must carefully analyze the words describing the situation. For example, if we say an unknown length is 2 more than twice another length, we must be able to translate the given information into an algebraic expression. In this case there are two lengths, with one length dependent on the other. If we let x represent the second length, then the first length is 2 more (+) than twice (2 ×) x, or 2 + 2x. 1 5.7.1: Building Functions from Context

2. Introduction, continued In this lesson, the problems will involve different contexts that quadratic equations describe. Notice that every problem will either require you to identify the two distinct linear expressions whose product will build the quadratic, or to identify the vertex of the parabola along with one other point on the parabola (which will allow you to build the quadratic in the vertex form of a quadratic equation). 2 5.7.1: Building Functions from Context

3. Key Concepts A quadratic expression is the product of two linear factors. The vertex of a parabola represents either the maximum or minimum y-value for the equation. The vertex of a parabola can be found using from the general form of the quadratic equation, where f(x) = ax 2 + bx + c. The vertex of a parabola can be found using (h, k) in vertex form, where f(x) = a(x – h) 2 + k. 3 5.7.1: Building Functions from Context

4. Key Concepts, continued The curve of a parabola is the graphical representation of the solution set for the equation of the parabola. A quadratic equation can be built using the vertex and any other point on the parabola. Concavity refers to the direction the parabola faces. The vertex of an upward-facing (or concave up) parabola will be at the bottom or minimum of the curve. Conversely, the vertex of a downward-facing (or concave down) parabola will be at the top or maximum of the parabola. 4 5.7.1: Building Functions from Context

5. Key Concepts, continued 5 5.7.1: Building Functions from Context Concave UpConcave Down The vertex is a minimum.The vertex is a maximum.

6. Key Concepts, continued The leading coefficient is the coefficient of the term with the highest power. In a quadratic equation, the leading coefficient is the number that is being multiplied by the x 2 term. The leading coefficient of the equation of a parabola determines its concavity. If the leading coefficient is positive, the parabola is concave up and the graph has a minimum. If the leading coefficient is negative, the parabola is concave down and the graph has a maximum. 6 5.7.1: Building Functions from Context

7. Common Errors/Misconceptions not realizing that when given a fixed perimeter, the corresponding area of the figure is constant thinking that a positive leading coefficient yields a maximum value instead of a minimum value mistakenly believing that adding linear expressions will build quadratics; they must be multiplied 7 5.7.1: Building Functions from Context

8. Guided Practice Example 1 A farmer is building a rectangular pen using 100 feet of electric fencing and the side of a barn. In addition to fencing, there will be a 4-foot gate also requiring the electric fencing on either side of the pen. The farmer wants to maximize the area of the pen. How long should he make each side of the fence in order to create the maximum area? 8 5.7.1: Building Functions from Context

9. Guided Practice: Example 1, continued 1.Write the expressions that describe the length of each side of the pen. Starting with one of the sides that is perpendicular to the barn, we can say the length is an unknown amount, x, plus the 4-foot gate, or x + 4. The length of the second side of the fence that is perpendicular to the barn will be the same length as the first. 9 5.7.1: Building Functions from Context

10. Guided Practice: Example 1, continued The side that is parallel to the barn is whatever amount of fence is left over after creating the two perpendicular sides. The total amount of fencing is 100 feet, and there are two sides of length (x + 4), so the length of the side that is parallel to the barn is 100 – 2(x + 4). Simplifying the expression, we get 100 – 2x – 8, or 92 – 2x. 10 5.7.1: Building Functions from Context

11. Guided Practice: Example 1, continued 2.Build the equation that describes the area of the pen. Remember that area equals length times width, or A = l w. Let l = x + 4 and w = 92 – 2x. 11 5.7.1: Building Functions from Context A(x) = (x + 4)(92 – 2x) Substitute values for length and width. A(x) = 92x – 2x 2 + 368 – 8xMultiply. A(x) = –2x 2 + 84x + 368Reorder and simplify.

12. Guided Practice: Example 1, continued 3.To find the maximum area, use the vertex. Note that the leading coefficient of the equation is negative. This means the graph of the equation will be a downward-facing parabola. Therefore, the vertex of the parabola will describe the greatest amount of area. A(x) = –2x 2 + 84x + 368 Given that the general form of a quadratic function is y = ax 2 + bx + c, we can determine that a = –2, b = 84, and c = 368. 12 5.7.1: Building Functions from Context

13. Guided Practice: Example 1, continued Find the x-coordinate of the vertex by substituting in a and b values from the quadratic function into the expression The x-coordinate of the vertex is 21. 13 5.7.1: Building Functions from Context

14. Guided Practice: Example 1, continued Find the y-coordinate of the vertex by substituting the x-coordinate from the vertex into the quadratic function. The maximum area of the pen is 1,250 ft 2. 14 5.7.1: Building Functions from Context A(x) = –2x 2 + 84x + 368 Quadratic function A(21) = –2(21) 2 + 84(21) + 368Substitute 21 for x. A(21) = –882 + 1764 + 368 Simplify, then solve. A(21) = 1250

15. Guided Practice: Example 1, continued 4.Finally, use the x-value from the vertex to find the lengths of each side of the pen. x = 21 Each side that is perpendicular to the barn is equal to x + 4. (x + 4) = (21 + 4) = 25 feet The side that is parallel to the barn is equal to 92 – 2x. (92 – 2x) = [92 – 2(21)] = 50 feet 15 5.7.1: Building Functions from Context ✔

16. Guided Practice: Example 1, continued 16 5.7.1: Building Functions from Context

17. Guided Practice Example 2 An amusement park has commissioned the design of a steel roller coaster with a drop section that is modeled by a parabola. Part of the roller coaster’s track will go through an underground tunnel. In this section, the roller coaster will dip 12 feet below ground level. The roller coaster will dip below ground level at a horizontal distance of 24 feet from the peak just before the drop and reemerge to ground level at a horizontal distance of 36 feet from the peak just before the drop. Find an equation of the parabola that describes the drop and the height of the roller coaster at the peak. 17 5.7.1: Building Functions from Context

18. Guided Practice: Example 2, continued 1.Derive the coordinates of the points on the parabola. Let the x-axis represent ground level and the y- intercept represent the peak of this section of the roller coaster. Since the horizontal distances are given as 24 feet and 36 feet, it can be determined that the roller coaster dips below ground level at (24, 0) and reemerges at (36, 0). 18 5.7.1: Building Functions from Context

19. Guided Practice: Example 2, continued 2.Establish the linear factors and the equation of the parabola. From the x-intercepts, the linear factors of the equation are (x – 24) and (x – 36). Therefore, the equation of the parabola is y = a(x – 24)(x – 36). 19 5.7.1: Building Functions from Context

20. Guided Practice: Example 2, continued 3.Find the vertex of the parabola. Expand the factored form of the equation of the parabola into the general form, y = ax 2 + bx + c. 20 5.7.1: Building Functions from Context y = a(x – 24)(x – 36)Equation of the parabola y = a(x 2 – 36x – 24x + 864)Multiply the factors. y = a(x 2 – 60x + 864)Simplify. y = ax 2 – 60ax + 864aDistribute a.

21. Guided Practice: Example 2, continued The x-coordinate for the vertex of a quadratic equation is Substitute the values from the quadratic function to determine the x-coordinate. The x-coordinate is 30. Combine this information with the given minimum of the drop, 12 feet below ground level, to find the vertex of the equation: (30, –12). 21 5.7.1: Building Functions from Context

22. Guided Practice: Example 2, continued 4.Find the value of a. Substitute the point (30, –12) into the factored form of the quadratic equation and solve for a. 22 5.7.1: Building Functions from Context y = a(x – 24)(x – 36) Factored form of the quadratic equation –12 = a(30 – 24)(30 – 36) Substitute 30 and –12 for x and y.

23. Guided Practice: Example 2, continued After solving for a, the equation of the parabola is 23 5.7.1: Building Functions from Context –12 = a(6)(–6)Simplify, then solve for a. –12 = –36a

24. Guided Practice: Example 2, continued 5.Find the height of the peak of this section of the roller coaster. The height of the peak is the roller coaster’s vertical distance above ground level. Since the peak occurs at the y-intercept, where the x-coordinate equals 0, set x = 0 in the quadratic equation and solve for y. 24 5.7.1: Building Functions from Context

25. Guided Practice: Example 2, continued The height of the roller coaster’s peak just before the drop is 288 feet. 25 5.7.1: Building Functions from Context ✔

26. Guided Practice: Example 2, continued 26 5.7.1: Building Functions from Context