UNIT 5 QUADRATIC FUNCTIONS Lesson 7: Building Functions Instruction

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1 Prerequisite Skills This lesson requires the use of the following skills: multiplying linear expressions factoring quadratic equations finding the value of a in the vertex form of a quadratic equation given the vertex and a point on the parabola finding the x- and y-coordinates of the vertex of a parabola 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. 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). 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. b The vertex of a parabola can be found using 2, a f b 2a 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. 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. U5-356

2 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. Concave Up Concave Down The vertex is a minimum. The vertex is a maximum. 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. 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 U5-357

3 Guided Practice 5.7. Example A farmer is building a rectangular pen using 00 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?. 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. 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 00 feet, and there are two sides of length (x + 4), so the length of the side that is parallel to the barn is 00 2(x + 4). Simplifying the expression, we get 00 2x 8, or 92 2x. 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. A(x) = (x + 4)(92 2x) A(x) = 92x 2x x A(x) = 2x x Substitute values for length and width. Multiply. Reorder and simplify. U5-358

4 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 x 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. Find the x-coordinate of the vertex by substituting in a and b values from the quadratic function into the expression b 2 a : b ( 84) = = 84 2a = 2 ( ) The x-coordinate of the vertex is 2. Find the y-coordinate of the vertex by substituting the x-coordinate from the vertex into the quadratic function. A(x) = 2x x Quadratic function A(2) = 2(2) (2) Substitute 2 for x. A(2) = A(2) = 250 The maximum area of the pen is,250 ft 2. Simplify, then solve. 4. Finally, use the x-value from the vertex to find the lengths of each side of the pen. x = 2 Each side that is perpendicular to the barn is equal to x + 4. (x + 4) = (2 + 4) = 25 feet The side that is parallel to the barn is equal to 92 2x. (92 2x) = [92 2(2)] = 50 feet U5-359

5 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 2 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.. 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). 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). 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. y = a(x 24)(x 36) y = a(x 2 36x 24x + 864) y = a(x 2 60x + 864) Equation of the parabola Multiply the factors. Simplify. y = ax 2 60ax + 864a Distribute a. The x-coordinate for the vertex of a quadratic equation is b 2 a. Substitute the values from the quadratic function to determine the x-coordinate. (continued) U5-360

6 b ( 60a) 30a = = = 30 2a 2( a) a The x-coordinate is 30. Combine this information with the given minimum of the drop, 2 feet below ground level, to find the vertex of the equation: (30, 2). 4. Find the value of a. Substitute the point (30, 2) into the factored form of the quadratic equation and solve for a. y = a(x 24)(x 36) Factored form of the quadratic equation 2 = a(30 24)(30 36) Substitute 30 and 2 for x and y. 2 = a(6)( 6) Simplify, then solve for a. 2 = 36a a = 3 After solving for a, the equation of the parabola is y= ( x 24)( x 36) 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. y = ( 0 24)( 0 36) 3 y = ( 24)( 36) 3 y = ( ) y = 288 The height of the roller coaster s peak just before the drop is 288 feet. U5-36

7 Example 3 A suspension bridge has two cables secured at either end of the span by two supporting towers. The cables are attached to the tops of the towers. In the section between the two towers, the cables form a parabolic curve. At their lowest point, the cables are 5 feet from the surface of the bridge. The towers are 400 feet apart, and the vertical distance from the surface of the bridge to the top of each tower is 45 feet. What is a quadratic equation that describes the curve of the cables between the towers? y x. Determine the vertex of the parabola. Since the cables are attached to the tops of the towers, this means the cables start 45 feet up from the surface of the bridge. The cables dip down to a height of 5 feet above the bridge. If we allow the y-axis to represent the midpoint between the towers, then we can say the vertex of the parabola is (0, 5). U5-362

8 2. Derive the equation of the parabola. Substitute the vertex (0, 5) into the general equation for the vertex form of a parabola, y = a(x h) 2 + k, so that h = 0 and k = 5. y = a(x h) 2 + k General equation y = a(x 0) Substitute 0 for h and 5 for k. y = ax Simplify. The equation of the parabola formed by the bridge cables is y = ax Find another point on the parabola. To solve for a, we need any other point on the parabola to substitute into the equation for x and y. In this case, we have been given the height of the supporting towers (45 feet) and their distance apart (400 feet). Since we let the axis of symmetry of the parabola be the y-axis, we can determine where the top of each tower lies on the parabola. The y-coordinate for the top of each tower is 45, the height. To find the x-coordinate, divide the horizontal distance between the towers, 400 feet, by = Since the y-axis is the axis of symmetry of the parabola, and each y-coordinate is positive, we can determine that one tower is in Quadrant I, (x, y). The other tower is in Quadrant II, ( x, y). Substitute the x- and y-values into these forms to determine where the tops of the towers are located. Tower in Quadrant I: (x, y) = (200, 45) Tower in Quadrant II: ( x, y) = ( 200, 45) The tops of the towers lie at ( 200, 45) and (200, 45). Therefore, we know the points ( 200, 45) and (200, 45) lie on the parabola. (continued) U5-363

9 ( 200, 45) (200, 45) y (0, 5) x 4. Solve for a. Take one of the known points that lies on the parabola and substitute the x- and y-values into the equation of the parabola. Either point will do. Let s use (200, 45). y = ax Equation of the parabola 45 = a(200) Substitute 200 for x and 45 for y. 45 = a(40,000) + 5 Simplify. 400 = a(40,000) Subtract 5 from both sides. 400 = a Divide both sides by 40, , 000 a = 00 Simplify. U5-364

10 5. Complete the equation of the curvature of the cable. Substitute for a in the equation of the parabola. 00 y = ax 2 + 5, where a = 00 2 y= x U5-365