 # Quadratic Functions and Their Graphs

## Presentation on theme: "Quadratic Functions and Their Graphs"— Presentation transcript:

Objectives Graphs of Quadratic Functions Min-Max Applications

The graph of any quadratic function is a parabola.
The vertex is the lowest point on the graph of a parabola that opens upward and the highest point on the graph of a parabola that opens downward.

The graph is symmetric with respect to the y-axis
The graph is symmetric with respect to the y-axis. In this case the y-axis is the axis of symmetry for the graph.

Example Use the graph of the quadratic function to identify the vertex, axis of symmetry, and whether the parabola opens upward or downward. a. b. Vertex (0, 2) Axis of symmetry: x = –2 Open: up Vertex (0, 4) Axis of symmetry: x = 0 Open: down

Example Find the vertex for the graph of Support your answer graphically. Solution a = 2 and b = 8 Substitute into the equation to find the y-value. The vertex is (2, 11), which is supported by the graph.

Example Identify the vertex, and the axis of symmetry on the graph, then graph. Solution Begin by making a table of values. Plot the points and sketch a smooth curve. The vertex is (0, –2) axis of symmetry x = 0 x f(x) = x2 – 2 3 7 2 2 1 1 3

Example Identify the vertex, and the axis of symmetry on the graph, then graph. Solution Begin by making a table of values. Plot the points and sketch a smooth curve. The vertex is (2, 0) axis of symmetry x = 2 x g(x) = (x – 2)2 4 1 1 2 3 4

Example Identify the vertex, and the axis of symmetry on the graph, then graph. Solution Begin by making a table of values. Plot the points and sketch a smooth curve. The vertex is (2, 0) axis of symmetry x = 2 x h(x) = x2 – 2x – 3 2 5 1 3 1 4 2 3 4

Example Find the maximum y-value of the graph of Solution The graph is a parabola that opens downward because a < 0. The highest point on the graph is the vertex. a = 1 and b = 2

Example A baseball is hit into the air and its height h in feet after t seconds can be calculated by a. What is the height of the baseball when it is hit? b. Determine the maximum height of the baseball. Solution a. The baseball is hit when t = 0. b. The graph opens downward because a < 0. The maximum height occurs at the vertex. a = –16 and b = 64.

Example (cont) The maximum height is 66 feet.

Basic Transformations of Graphs
The graph of y = ax2, a > 0. As a increases, the resulting parabola becomes narrower. When a > 0, the graph of y = ax2 never lies below the x-axis.

Example Compare the graph of g(x) = –4x2 to the graph of f(x) = x2. Then graph both functions on the same coordinate axes. Solution Both graphs are parabolas. The graph of g opens downward and is narrower than the graph of f.

Parabolas and Modeling
Section 11.2 Parabolas and Modeling Copyright © 2013, 2009, and 2005 Pearson Education, Inc.

Objectives Vertical and Horizontal Translations Vertex Form

Vertical and Horizontal Translations
The graph of y = x2 is a parabola opening upward with vertex (0, 0). All three graphs have the same shape. y = x2 y = x2 + 1 shifted upward 1 unit y = x2 – 2 shifted downward 2 units Such shifts are called translations because they do not change the shape of the graph only its position

Vertical and Horizontal Translations
The graph of y = x2 is a parabola opening upward with vertex (0, 0). y = x2 y = (x – 1)2 Horizontal shift to the right 1 unit

Vertical and Horizontal Translations
The graph of y = x2 is a parabola opening upward with vertex (0, 0). y = x2 y = (x + 2)2 Horizontal shift to the left 2 units

Example Sketch the graph of the equation and identify the vertex. Solution The graph is similar to y = x2 except it has been translated 3 units down. The vertex is (0, 3).

Example Sketch the graph of the equation and identify the vertex. Solution The graph is similar to y = x2 except it has been translated left 4 units. The vertex is (4, 0).

Example Sketch the graph of the equation and identify the vertex. Solution The graph is similar to y = x2 except it has been translated down 2 units and right 1 unit. The vertex is (1, 2).

Example Compare the graph of y = f(x) to the graph of y = x2. Then sketch a graph of y = f(x) and y = x2 in the same xy-plane. Solution The graph is translated to the right 2 units and upward 3 units. The vertex for f(x) is (2, 3) and the vertex of y = x2 is (0, 0). The graph opens upward and is wider.

Example Write the vertex form of the parabola with a = 3 and vertex (2, 1). Then express the equation in the form y = ax2 + bx + c. Solution The vertex form of the parabola is where the vertex is (h, k). a = 3, h = 2 and k = 1 To write the equation in y = ax2 + bx + c, do the following:

Example Write each equation in vertex form. Identify the vertex. a. b. Solution a. Because , add and subtract 16 on the right.

Example (cont) b. This equation is slightly different because the leading coefficient is 2 rather than 1. Start by factoring 2 from the first two terms on the right side.

Objectives Basics of Quadratic Equations The Square Root Property
Completing the Square Solving an Equation for a Variable Applications of Quadratic Equations

Any quadratic function f can be represented by f(x) = ax2 + bx + c with a  0. Examples:

Basics The different types of solutions to a quadratic equation.

Example Solve each quadratic equation. Support your results numerically and graphically. a. b. c. Solution a. Symbolic: Numerical: Graphical: x y 1 5 2 1 The equation has no real solutions because x2 ≥ 0 for all real numbers x.

Example (cont) b. The equation has one real solution. x y 5 4 4 1 3
2 1 b. The equation has one real solution.

Example (cont) c. The equation has two real solutions. x y 4 2 8 1
2 8 1 9 2 The equation has two real solutions.

The Square Root Property
The square root property is used to solve quadratic equations that have no x-terms.

Example Solve each equation. a. b. c. Solution a. b. c.

Real World Connection If an object is dropped from a height of h feet, its distance d above the ground after t seconds is given by

Example A toy falls 40 feet from a window. How long does the toy take to hit the ground? Solution

Example Find the term that should be added to to form a perfect square trinomial. Solution Coefficient of x-term is –8, so we let b = –8. To complete the square we divide by 2 and then square the result.

Example Solve the equation Solution Write the equation in x2 + bx = d form.

Example Solve the equation Solution Write the equation in x2 + bx = d form.

Example Solve the equation for the specified variable. Solution

Example Use of the Internet in Western Europe has increased dramatically. The figure shows a scatter plot of online users in Western Europe, together with a graph of a function f that models the data. The function f is given by: where the output is in millions of users. In this formula x = 6 corresponds to 1996, x = 7 to 1997, and so on, until x = 12 represents a. Evaluate f(10) and interpret the result. b. Graph f and estimate the year when the number of Internet users reached 85 million. c. Solve part (b) numerically.

Example (cont) Solution a. Evaluate f(10) and interpret the result. Because x = 10 corresponds to 2000, there were about 51.4 million users in 2000.

Example (cont) Solution b. Graph f and estimate the year when the number of Internet users reached 85 million. c. Solve part (b) numerically.

Objectives Solving Quadratic Equations The Discriminant

QUADRATIC FORMULA The solutions to ax2 + bx + c = 0 with a ≠ 0 are given by

Example Solve the equation 4x2 + 3x – 8 = 0. Support your results graphically. Solution Symbolic Solution Let a = 4, b = 3 and c = − 8. or or

Example (cont) 4x2 + 3x – 8 = 0 Graphical Solution

Example Solve the equation 3x2 − 6x + 3 = 0. Support your result graphically. Solution Let a = 3, b = −6 and c = 3.

Example Solve the equation 2x2 + 4x + 5 = 0. Support your result graphically. Solution Let a = 2, b = 4 and c = 5. There are no real solutions for this equation because is not a real number.

EQUATIONS To determine the number of solutions to the quadratic equation ax2 + bx + c = 0, evaluate the discriminant b2 – 4ac. 1. If b2 – 4ac > 0, there are two real solutions. 2. If b2 – 4ac = 0, there is one real solution. 3. If b2 – 4ac < 0, there are no real solutions; there are two complex solutions.

Example Use the discriminant to determine the number of solutions to −2x2 + 5x = 3. Then solve the equation using the quadratic formula. Solution −2x2 + 5x − 3 = 0 Let a = −2, b = 5 and c = −3. b2 – 4ac = (5)2 – 4(−2)(−3) = 1 or Thus, there are two solutions.

THE EQUATION x2 + k = 0 If k > 0, the solution to x2 + k = 0 are given by

Example Solve x = 0. Solution The solutions are

Example Solve 3x2 – 7x + 5 = 0. Write your answer in standard form: a + bi. Solution Let a = 3, b = −7 and c = 5. and

Example Solve Write your answer in standard form: a + bi. Solution Begin by adding 2x to each side of the equation and then multiply by 5 to clear fractions. Let a = −2, b = 10 and c = −15.

Example (cont) Let a = −2, b = 10 and c = −15.

Example Solve by completing the square. Solution After applying the distributive property, the equation becomes Since b = −4 ,add to each side of the equation. The solutions are 2 + i and 2 − i.