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Algebra 2 Standard Form of a Quadratic Function Lesson 4-2 Part 1

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Goals Goal To graph quadratic functions written in standard form. Rubric Level 1 – Know the goals. Level 2 – Fully understand the goals. Level 3 – Use the goals to solve simple problems. Level 4 – Use the goals to solve more advanced problems. Level 5 – Adapts and applies the goals to different and more complex problems.

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Vocabulary Standard Form

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Essential Question Big Idea: Function and Equivalence Why is the standard form of a quadratic function useful?

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Quadratic Function Standard Form Another useful form of writing quadratic functions is the standard form. The standard form of a quadratic function is f(x)= ax 2 + bx + c, where a ≠ 0. The coefficients a, b, and c can show properties of the graph of the function.

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Graphing a Quadratic Function in Standard Form Two of the defining characteristics of a quadratic function are its vertex and axis of symmetry. Therefore, when graphing a quadratic function, the first thing you want to calculate are the vertex and the axis of symmetry. The vertex and axis of symmetry are related. The vertex is on the axis of symmetry and you can use the equation of the axis of symmetry to find the x-coordinate of the vertex.

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Calculating the Properties of a Quadratic Function in Standard FORM 1.Axis of Symmetry 2.The Vertex 3. The y-Intercept

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The axis of symmetry is the line through the vertex of a parabola that divides the parabola into two congruent halves. Review Axis of Symmetry

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You can use a formula to calculate the axis of symmetry. The formula works for all quadratic functions. Standard Form Calculating the Axis of Symmetry

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Procedure: 1.Using the standard form of the quadratic function, y = ax 2 + bx + c, find the values of a and b. 2.Use the formula,. 3.Write the axis of symmetry as an equation. Standard Form Calculating the Axis of Symmetry

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Find the axis of symmetry of the graph of y = – 3x 2 + 10x + 9. Step 1. Find the values of a and b. y = –3x 2 + 10x + 9 a = –3, b = 10 Step 2. Use the formula. The axis of symmetry is Example: Step 3. Write as an equation.

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Find the axis of symmetry of the graph of y = 2x 2 + x + 3. Step 1. Find the values of a and b. y = 2x 2 + 1x + 3 a = 2, b = 1 Step 2. Use the formula. The axis of symmetry is. Your Turn: Step 3. Write as an equation.

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The graph of a quadratic function is a parabola. The vertex is the turning point of the parabola, the lowest (minimum) or highest (maximum) point on the curve. Axis of symmetry Vertex is the lowest point Vertex is the highest point Opens up a > 0 Opens down a < 0 Review - Vertex

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Once you have found the axis of symmetry, you can use it to identify the vertex. Standard Form Calculating the Vertex

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Find the vertex. y = –3x 2 + 6x – 7 Step 1 Find the x-coordinate of the vertex. a = –3, b = 10 Identify a and b. Substitute –3 for a and 6 for b. The x-coordinate of the vertex is 1. Example:

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Step 2 Find the corresponding y-coordinate. y = –3x 2 + 6x – 7 = –3(1) 2 + 6(1) – 7 = –3 + 6 – 7 = –4 Use the function rule. Substitute 1 for x. Step 3 Write the ordered pair. The vertex is (1, –4). y = –3x 2 + 6x – 7 Example: Continued The x-coordinate of the vertex is 1.

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Find the vertex. y = x 2 – 4x – 10 Step 1 Find the x-coordinate of the vertex. a = 1, b = –4 Identify a and b. Substitute 1 for a and –4 for b. The x-coordinate of the vertex is 2. Your Turn:

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Step 2 Find the corresponding y-coordinate. y = x 2 – 4x – 10 = (2) 2 – 4(2) – 10 = 4 – 8 – 10 = –14 Use the function rule. Substitute 2 for x. Step 3 Write the ordered pair. The vertex is (2, –14). y = x 2 – 4x – 10 The x-coordinate of the vertex is 2. Continued

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Recall that a y-intercept is the y-coordinate of the point where a graph intersects the y-axis. The x-coordinate of this point is always 0. For a quadratic function written in the form y = ax 2 + bx + c, when x = 0, y = c. So the y-intercept of a quadratic function is (0, c). Standard Form Finding the y-intercept

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These properties can be generalized to help you graph quadratic functions. Properties Quadratic Function in Standard Form

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Standard Form Graphing Quadratic Functions Procedure: 1.Find the axis of symmetry,. 2.Find the vertex, (h, k). 3.Find the y-intercept. 4.Find two more points on the same side of the axis of symmetry as the point containing the y-intercept. 5.Graph the axis of symmetry, the vertex, the point containing the y-intercept, and two other points. 6. Reflect the points across the axis of symmetry. Connect the points with a smooth curve. Confirm the direction the graph opens with the sign of a.

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Graph f(x) = 3x 2 – 6x + 1. Step 1 Find the axis of symmetry. = 1 The axis of symmetry is x = 1. Simplify. Use x =. Substitute 3 for a and –6 for b. Step 2 Find the vertex. y = 3x 2 – 6x + 1 = 3(1) 2 – 6(1) + 1 = 3 – 6 + 1 = –2= –2 The vertex is (1, –2). The x-coordinate of the vertex is 1. Substitute 1 for x. Simplify. The y-coordinate is –2. Example:

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Step 3 Find the y-intercept. y = 3x 2 – 6x + 1 The y-intercept is 1; the graph passes through (0, 1). Identify c. Example: Continued f(x) = 3x 2 – 6x + 1.

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Step 4 Find two more points on the same side of the axis of symmetry as the point containing the y-intercept. Since the axis of symmetry is x = 1, choose x-values less than 1. Let x = –1. y = 3(–1) 2 – 6(–1) + 1 = 3 + 6 + 1 = 10 Let x = –2. y = 3(–2) 2 – 6(–2) + 1 = 12 + 12 + 1 = 25 Substitute x-coordinates. Simplify. Two other points are (–1, 10) and (–2, 25). Example: Continued f(x) = 3x 2 – 6x + 1.

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Step 5 Graph the axis of symmetry, the vertex, the point containing the y-intercept, and two other points. Step 6 Reflect the points across the axis of symmetry. Connect the points with a smooth curve. Confirm a > 0, opens up. x = 1 (–2, 25) (–1, 10) (0, 1) (1, –2) x = 1 (–1, 10) (0, 1) (1, –2) (–2, 25) Example: Continued f(x) = 3x 2 – 6x + 1.

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Because a parabola is symmetrical, each point is the same number of units away from the axis of symmetry as its reflected point. Helpful Hint

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Graph the quadratic function. f(x) = 2x 2 + 6x + 2 Step 1 Find the axis of symmetry. Simplify. Use x =. Substitute 2 for a and 6 for b. The axis of symmetry is x. Example:

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Step 2 Find the vertex. y = 2x 2 + 6x + 2 Simplify. = 4 – 9 + 2= –2= –2 The x-coordinate of the vertex is. Substitute for x. The y-coordinate is. The vertex is. Continued f(x) = 2x 2 + 6x + 2

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Step 3 Find the y-intercept. y = 2x 2 + 6x + 2 The y-intercept is 2; the graph passes through (0, 2). Identify c. Continued f(x) = 2x 2 + 6x + 2

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Step 4 Find two more points on the same side of the axis of symmetry as the point containing the y-intercept. Let x = –1 y = 2(–1) 2 + 6(–1) + 1 = 2 – 6 + 2 = –2 Let x = 1 y = 2(1) 2 + 6(1) + 2 = 2 + 6 + 2 = 10 Substitute x-coordinates. Simplify. Two other points are (–1, –2) and (1, 10). Since the axis of symmetry is x = –1, choose x values greater than –1. Continued f(x) = 2x 2 + 6x + 2

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Step 5 Graph the axis of symmetry, the vertex, the point containing the y-intercept, and two other points. Step 6 Reflect the points across the axis of symmetry. Connect the points with a smooth curve. Confirm a > 0, opens up. (–1, –2) (1, 10) (–1, –2) (1, 10) Continued f(x) = 2x 2 + 6x + 2

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Graph the quadratic function. y + 6x = x 2 + 9 Step 1 Find the axis of symmetry. Simplify. Use x =. Substitute 1 for a and –6 for b. The axis of symmetry is x = 3. = 3= 3 f(x) = x 2 – 6x + 9 Rewrite in standard form. Your Turn:

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Step 2 Find the vertex. Simplify. = 9 – 18 + 9 = 0= 0 The vertex is (3, 0). The x-coordinate of the vertex is 3. Substitute 3 for x. The y-coordinate is 0.. y = x 2 – 6x + 9 y = 3 2 – 6(3) + 9 Continued f(x) = x 2 – 6x + 9

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Step 3 Find the y-intercept. y = x 2 – 6x + 9 The y-intercept is 9; the graph passes through (0, 9). Identify c. Continued f(x) = x 2 – 6x + 9

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Step 4 Find two more points on the same side of the axis of symmetry as the point containing the y- intercept. Since the axis of symmetry is x = 3, choose x-values less than 3. Let x = 2 y = 1(2) 2 – 6(2) + 9 = 4 – 12 + 9 = 1 = 1 Let x = 1 y = 1(1) 2 – 6(1) + 9 = 1 – 6 + 9 = 4 = 4 Substitute x-coordinates. Simplify. Two other points are (2, 1) and (1, 4). Continued f(x) = x 2 – 6x + 9

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Step 5 Graph the axis of symmetry, the vertex, the point containing the y-intercept, and two other points. Step 6 Reflect the points across the axis of symmetry. Connect the points with a smooth curve. Confirm a > 0, opens up. x = 3 (3, 0) (0, 9) (2, 1) (1, 4) (0, 9) (1, 4) (2, 1) x = 3 (3, 0) Continued f(x) = x 2 – 6x + 9

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Graph the function f(x) = –x 2 – 2x + 3. Step 1 Find the axis of symmetry. The axis of symmetry is the line x = –1. Substitute –2 for b and –1 for a. The axis of symmetry is given by. Your Turn:

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Step 2 Find the vertex. The vertex lies on the axis of symmetry, so the x-coordinate is –1. The y-coordinate is the value of the function at this x-value, or f(–1). f(–1) = –(–1) 2 – 2(–1) + 3 = 4 The vertex is (–1, 4). Step 3 Find the y-intercept. Because c = 3, the y-intercept is (0, 3). f(x) = –x 2 – 2x + 3 Continued

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Step 4 Find two more points on the same side of the axis of symmetry as the point containing the y-intercept. Let x = 1 y = -(1) 2 - 2(1) + 3 = -1 - 2 + 3 = 0 = 0 Let x = 2 y = -(2) 2 - 2(2) + 3 = -4 - 4 + 3 = -5 Substitute x-coordinates. Simplify. Two other points are (1, 0) and (2, -5). Since the axis of symmetry is x = -1, choose x values greater than -1. Continued f(x) = –x 2 – 2x + 3

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Step 5 Graph the axis of symmetry (x = -1), the vertex (-1, 4), the point containing the y-intercept (0, 3), and two other points (1, 0) & (2, -5). Step 6 Reflect the points across the axis of symmetry. Connect the points with a smooth curve. Confirm a < 0, opens down. Continued f(x) = –x 2 – 2x + 3

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Graph the function f(x) = 2x 2 – 4x + 5. Step 1 Find the axis of symmetry. The axis of symmetry is the line x = 1. Substitute –4 for b and 2 for a. The axis of symmetry is given by. Your Turn:

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f(x) = 2x 2 – 4x + 5 Step 2 Find the vertex. The vertex lies on the axis of symmetry, so the x-coordinate is 1. The y-coordinate is the value of the function at this x-value, or f(1). f(1) = 2(1) 2 – 4(1) + 5 = 3 The vertex is (1, 3). Step 3 Find the y-intercept. Because c = 5, the y-intercept is (0, 5). Continued

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Step 4 Find two more points on the same side of the axis of symmetry as the point containing the y-intercept. Let x = -1 y = 2(-1) 2 - 4(-1) + 5 = 2 + 4 + 5 = 11 Let x = -2 y = 2(-2) 2 - 4(-2) + 5 = 8 + 8 + 5 = 21 Substitute x-coordinates. Simplify. Two other points are (-1, 11) and (-2, 21). Since the axis of symmetry is x = 1, choose x values less than 1. Continued f(x) = 2x 2 – 4x + 5

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Step 5 Graph the axis of symmetry (x = 1), the vertex (1, 3), the point containing the y-intercept (0, 5), and two other points (-1, 11) & (-2, 21). Step 6 Reflect the points across the axis of symmetry. Connect the points with a smooth curve. Confirm a > 0, opens up. Continued f(x) = 2x 2 – 4x + 5

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The axis of symmetry is the line x = –1. Substitute –4 for b and –2 for a. Graph the function f(x)= –2x 2 – 4x Step 1 The axis of symmetry is given by. Your Turn:

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Step 2 The vertex lies on the axis of symmetry, so the x-coordinate is –1. The y-coordinate is the value of the function at this x-value, or f(–1). f(–1) = –2(–1) 2 – 4(–1) = 2 The vertex is (–1, 2). Step 3 Because c is 0, the y-intercept is (0, 0). f(x)= –2x 2 – 4x Continued

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Step 4 Find two more points on the same side of the axis of symmetry as the point containing the y-intercept. Let x = 1 y = -2(1) 2 - 4(1) = -2 - 4 = -6 Let x = 2 y = -2(2) 2 - 4(2) = -8 - 8 = -16 Substitute x-coordinates. Simplify. Two other points are (1, -6) and (2, -16). Since the axis of symmetry is x = -1, choose x values greater than -1. Continued f(x)= –2x 2 – 4x

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Step 5 Graph the axis of symmetry (x = -1), the vertex (-1, 2), the point containing the y-intercept (0, 0), and two other points (1, -6) & (2, -16). Step 6 Reflect the points across the axis of symmetry. Connect the points with a smooth curve. Confirm a < 0, opens down. Continued f(x)= –2x 2 – 4x

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Essential Question Big Idea: Function and Equivalence Why is the standard form of a quadratic function useful? Like vertex form, it is still possible to find information about the graph of a quadratic function. However, standard form is easier to enter into a graphing calculator.

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Assignment Section 4-2 pt 1, Pg. 214 – 215; #1 – 4 all, 6 – 22 even.

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