Presentation on theme: "Quadratic Functions and Equations"— Presentation transcript:
1 Quadratic Functions and Equations Chapter 3Quadratic Functions and Equations
2 Quadratic Functions and Models 3.1Learn basic concepts about quadratic functions and their graphs.Complete the square and apply the vertex formula.Graph a quadratic function by hand.Solve applications and model data.Use quadratic regression to model data (optional)
3 Basic ConceptsRecall that a linear function can be written as f(x) = ax + b (or f(x) = mx + b). The formula for a quadratic function is different from that of a linear function because it contains an x2 term. f(x) = 3x2 + 3x + 5 g(x) = 5 x2
4 Quadratic FunctionLet a, b, and c be constants with a ≠ 0. A function represented by f(x) = ax2 + bx + c is a quadratic function.
5 Quadratic FunctionThe graph of a quadratic function is a parabola—a U shaped graph that opens either upward or downward. A parabola opens upward if a is positive and opens downward if a is negative. The highest point on a parabola that opens downward and the lowest point on a parabola that opens upward is called the vertex.
6 Quadratic FunctionThe vertical line passing through the vertex is called the axis of symmetry. The leading coefficient a controls the width of the parabola. Larger values of |a| result in a narrower parabola, and smaller values of |a| result in a wider parabola.
8 Example: Analyzing graphs of quadratic functions Use the graph of the quadratic function to determine the sign of the leading coefficient, its vertex, and the equation of the axis of symmetry. Give intervals where the function is increasing and where it is decreasing. Give the domain and range.SolutionLeading coefficient: Opens downward, so a is negative.Vertex: The vertex is (–2, 5).Axis of symmetry: Vertical line through the vertex with equation x = –2.
9 Example: Analyzing graphs of quadratic functions Increasing for x ≤ –2.Decreasing for x ≥ –2.Domain: All real numbers.Range:
10 Vertex FormThe parabolic graph of f(x) = a(x – h)2 + k with a ≠ 0 has vertex (h, k). Its graph opens upward when a > 0 and opens downward when a < 0.
11 Completing the SquareWe can convert the general form f(x) = ax2 + bx + c to vertex form by completing the square. If a quadratic expression can be written as then it is a perfect square trinomial and can be factored as
12 Example: Converting to Vertex Form Write the formula f(x) = x2 + 6x – 3 in vertex form by completing the square. SolutionGiven formulaSubtract 3 from each side.Let k = 10; add (10/2)2 = 25.Factor perfect square trinomial.Subtract 12.Required form.
13 Vertex Formula The vertex of the graph of f(x) = ax2 + bx + c with a ≠ 0 is the point
14 Example: ConvertingUse the vertex formula to write f(x) = 3x2 + 12x + 7 in vertex form. Solution Vertex: (–2, –5)
15 Graphing Quadratic Functions When sketching a parabola, it is important to determine the vertex, the axis of symmetry, and whether the parabola opens upward or downward.
16 Example: Graphing quadratic functions by hand Graph the quadratic function Solution The formula is not in vertex form, but we can find the vertex. The y-coordinate of the vertex is: The vertex is at (–1, 5/2). The axis of symmetry is x = –1, and the parabola opens downward because a = –1/2 is negative
17 Example: Graphing quadratic functions by hand Table of ValuesGraph:
18 Applications and Models Sometimes when a quadratic function f is used in applications, the vertex provides important information. The reason is that the y-coordinate of the vertex is the minimum value of f(x) when its graph opens upward and is the maximum value of f(x) when its graph opens downward.
19 Example: Maximizing area A rancher is fencing a rectangular area for cattle using the straight portion of a river as one side of the rectangle. If the rancher has 2400 feet of fence, find the dimensions of the rectangle that give the maximum area for the cattle. Solution Let W be the width and L be the length of the rectangle. Because the 2400-foot fence does not go along the river, it follows that W + L + W = 2400 or L = 2400 – 2W
20 Example: Maximizing area Area of the rectangle equals length times width. This is a parabola that opens downward, and by the vertex formula, the maximum area occurs when
21 Example: Maximizing area The corresponding length is L = 2400 – 2W = 2400 – 2(600) = 1200 feet. The dimensions are 600 feet by 1200 feet.
22 Applications and Models Another application of quadratic functions occurs in projectile motion, such as when a baseball is hit up in the air. If air resistance is ignored, then the formula s(t) = –16t2 + v0t + h0 calculates the height s of the object above the ground in feet after t seconds. In this formula h0 represents the initial height of the object in feet and v0 represents its initial vertical velocity in feet per second. If the initial velocity is upward, then v0 > 0 and if the initial velocity is downward, then v0 < 0.
23 Example: ModelingA baseball is hit straight up with an initial velocity of v0 = 80 feet per second (or about 55 miles per hour) and leaves the bat with an initial height of h0 = 3 feet, a) Write a formula s(t) that models the height of the baseball after t seconds.b) How high is the baseball after 2 seconds?c) Find the maximum height of the baseball. Support your answer graphically.
24 Example: Modelinga) b) Baseball is 99 feet high after 2 seconds. c) Because a is negative, the vertex is the highest point on the graph, with a t-coordinate of
25 Example: Modeling c) The y-coordinate is: The vertex is at (2.5, 103). The maximum height of the baseball is 103 feet after 2.5 seconds.The graph supports this answer.