 ## Presentation on theme: "Graphing Quadratic Functions"— Presentation transcript:

Key Concepts The graph of a quadratic is U-shaped and called a parabola. The extremum of a graph is the function value that achieves either a maximum or a minimum. The maximum is the largest y-value of a quadratic, and the minimum is the smallest y-value of a quadratic. 5.6.1: Graphing Quadratic Functions

Key Concepts, continued
If a > 0, the parabola is concave up and the quadratic has a minimum. If a < 0, the parabola is concave down and the quadratic has a maximum. The extreme values of a quadratic occur at the vertex, the point at which the curve changes direction. If the quadratic equation is given in standard form, the vertex can be found by identifying the x-coordinate of the vertex using 5.6.1: Graphing Quadratic Functions

Key Concepts, continued
Substitute the value of the x-coordinate into the quadratic equation to find the y-coordinate of the vertex. The vertex of a quadratic function is The vertex form of a quadratic function is f(x) = a(x – h)2 + k, where the coordinate pair (h, k) is the location of the vertex. 5.6.1: Graphing Quadratic Functions

Intercepts The intercept of a graph is the point at which a line intercepts the x- or y-axis. The y-intercept of a function is the point at which the graph crosses the y-axis. This occurs when x = 0. The y-intercept is written as (0, y). The x-intercepts of a function are the points at which the graph crosses the x-axis. This occurs when y = 0. The x-intercept is written as (x, 0). 5.6.1: Graphing Quadratic Functions

Zeros The zeros of a function are the x-values for which the function value is 0. The intercept form of the quadratic function, written as f(x) = a(x – p)(x – q), where p and q are the zeros of the function, can be used to identify the x-intercepts. Set the factored form equal to 0. Then set each factor equal to 0. As long as the coefficients of x are 1, the x-intercepts are located at (r, 0) and (s, 0). These values for x are the roots, or the solutions, of the quadratic equation. 5.6.1: Graphing Quadratic Functions

Symmetry Parabolas are symmetrical; that is, they have two identical parts when rotated around a point or reflected over a line. This line is the axis of symmetry, the line through the vertex of a parabola about which the parabola is symmetric. The equation of the axis of symmetry is Symmetry can be used to find the vertex of a parabola if the vertex is not known. 5.6.1: Graphing Quadratic Functions

X-coordinate of the vertex
If you know the x-intercepts of the graph or any two points on the graph with the same y-value, the x-coordinate of the vertex is the point halfway between the values of the x-coordinates. For x-intercepts (r, 0) and (s, 0), the x-coordinate of the vertex is 5.6.1: Graphing Quadratic Functions

Practice # 1 Given the function f(x) = –2x2 + 16x – 30, identify the key features of the graph: the extremum, vertex, x-intercept(s), and y-intercept. Then sketch the graph. 5.6.1: Graphing Quadratic Functions

Determine the extremum of the graph
The extreme value is a minimum when a > 0. It is a maximum when a < 0. Because a = –2, the graph opens downward and the quadratic has a maximum. 5.6.1: Graphing Quadratic Functions

Determine the vertex of the graph
The maximum value occurs at the vertex. The vertex is of the form Use the original equation f(x) = –2x2 + 16x – 30 to find the values of a and b in order to find the x-value of the vertex. 5.6.1: Graphing Quadratic Functions

continued The x-coordinate of the vertex is 4. Formula to find the
x-coordinate of the vertex of a quadratic Substitute –2 for a and 16 for b. x = 4 Simplify. 5.6.1: Graphing Quadratic Functions

continued Substitute 4 into the original equation to find the y-coordinate. The y-coordinate of the vertex is 2. The vertex is located at (4, 2). f(x) = –2x2 + 16x – 30 Original equation f(4) = –2(4)2 + 16(4) – 30 Substitute 4 for x. f(4) = 2 Simplify. 5.6.1: Graphing Quadratic Functions

Determine the x-intercept(s) of the graph
Since the vertex is above the x-axis and the graph opens downward, there will be two x-intercepts. Factor the quadratic and set each factor equal to 0. 5.6.1: Graphing Quadratic Functions

continued f(x) = –2x2 + 16x – 30 Original equation
Factor out the greatest common factor. f(x) = –2(x – 3)(x – 5) Factor the trinomial. 0 = –2(x – 3)(x – 5) Set the factored form equal to 0 to find the intercepts. x – 3 = 0 or x – 5 = 0 Set each factor equal to 0 and solve for x. x = 3 or x = 5 Simplify. The x-intercepts are (3, 0) and (5, 0). 5.6.1: Graphing Quadratic Functions

Determine the y-intercept of the graph
The y-intercept occurs when x = 0. Substitute 0 for x in the original equation. The y-intercept is (0, –30). When the quadratic equation is written in standard form, the y-intercept is c. f(x) = –2x2 + 16x – 30 Original equation f(0) = –2(0)2 + 16(0) – 30 Substitute 0 for x. f(0) = –30 Simplify. 5.6.1: Graphing Quadratic Functions

Graph the function Use symmetry to identify additional points on the graph. The axis of symmetry goes through the vertex, so the axis of symmetry is x = 4. For each point to the left of the axis of symmetry, there is another point the same distance on the right side of the axis and vice versa. The point (0, –30) is on the graph, and 0 is 4 units to the left of the axis of symmetry. 5.6.1: Graphing Quadratic Functions

continued The point that is 4 units to the right of the axis is 8, so the point (8, –30) is also on the graph. Determine two additional points on the graph. Choose an x-value to the left or right of the vertex and find the corresponding y-value. f(x) = –2x2 + 16x – 30 Original equation f(1) = –2(1)2 + 16(1) – 30 Substitute 1 for x. f(1) = –16 Simplify. 5.6.1: Graphing Quadratic Functions

continued An additional point is (1, –16).
(1, –16) is 3 units to the left of the axis of symmetry. The point that is 3 units to the right of the axis is 7, so the point (7, –16) is also on the graph. Plot the points and join them with a smooth curve. 5.6.1: Graphing Quadratic Functions

Graph the function 5.6.1: Graphing Quadratic Functions

Your turn: Given the function f(x) = –2(x + 1)(x + 5), identify the key features of its graph: the extremum, vertex, x-intercept(s), and y-intercept. Then sketch the graph. 5.6.1: Graphing Quadratic Functions

Thanks for Watching! Ms. Dambreville