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Warm-up

Warm-up. Factor the following x 3 – 3x 2 – 28x 3x 2 – x – 4 16x 4 – 9y 2 x 3 + x 2 – 9x - 9. Section 2-1. Graphing Quadratic Functions. Objectives. I can graph quadratic functions in vertex form. I can convert to vertex format by completing the square

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Warm-up

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  1. Warm-up • Factor the following • x3 – 3x2 – 28x • 3x2 – x – 4 • 16x4 – 9y2 • x3 + x2 – 9x - 9

  2. Section 2-1 Graphing Quadratic Functions

  3. Objectives • I can graph quadratic functions in vertex form. • I can convert to vertex format by completing the square • I can find the equation of a quadratic given the vertex point and another point.

  4. Applications • Many applications that are used to model consumer behavior. • Example: Revenue generated from manufacturing handheld video games

  5. y vertex x Let a, b, and c be real numbers a 0. The function f(x) = ax2 + bx + c is called a quadratic function. Quadratic function The graph of a quadratic function is a parabola. Every parabola is symmetrical about a line called the axis (of symmetry). The intersection point of the parabola and the axis is called the vertex of the parabola. f(x) = ax2 + bx + c axis

  6. Quadratic Key Terms • Vertex: The vertex is the point of intersection between the Axis of Symmetry and the parabola. • Axis of Symmetry: This is the straight line that cuts the parabola into mirror images. • Solutions (Also called Roots): These are the x-coordinates of the points where the parabola intersects the x-axis.

  7. Two Real Solutions

  8. One Real Solution

  9. No Real Solutions

  10. y=ax2+bx+c • One key point to graph is always the y-intercept. This is a point that has the x value equal to zero (0, b) • If we let “x” be zero if the equation, then we see that the y-intercept is just the value of “c” in the equation. • Example: y = 2x2 – 3x + 4 • The y-intercept is (0, 4)

  11. Solutions versus Intercepts • Lets look at the differences in these vocabulary terms: • Solutions, roots, zeros, x-intercepts • If the problem asks you to find solutions or roots your answer should be in the format x = ?? Or {2, 3, 5}. We just list the x-value • If the problem asks you to find zeros or x-intercepts, your answer should be in ordered pair format: (2, 0) (-3, 0)

  12. y x vertex minimum y x vertex maximum Leading Coefficient The leading coefficient of ax2 + bx + c is a. a > 0 opens upward When the leading coefficientis positive, the parabola opens upward and the vertex is a minimum. f(x) = ax2 + bx + c When the leading coefficient is negative, the parabola opens downwardand the vertex is a maximum. f(x) = ax2 + bx + c a < 0 opens downward

  13. Example: Compare the graphs of , and y 5 x -5 5 The simplest quadratic functions are of the form f(x) = ax2 (a  0) Simple Quadratic Functions These are most easily graphed by comparing them with the graph ofy = x2.

  14. Standard Vertex Format • The format for any parabola with an equation ax2 + bx + c = 0 can be written in the following standard vertex format: • y = a(x – h)2 + k • Where the Vertex is (h,k) and the Axis of Symmetry is x = h • If a > o, then the parabola opens upward • If a < o, then the parabola opens downward

  15. Example 2: y = x2 – 4x + 5 • y = x2 – 4x + 5 (Now complete the square) • y = (x2 – 4x + ___) + 5 - _____ • -4/2 = -2 • (-2)2 = 4 • y = (x2 – 4x + 4) + 5 – 4 • y = (x – 2)2 + 1 • Vertex is (2, 1), Axis of Sym: x = 2, a > 0, so parabola turns upward

  16. Example 3: y = -5x2 + 80x - 319 • y = -5x2 + 80x - 319 (1st factor –5 from 1st two terms) • y = -5(x2 – 16x) - 319 (Now complete square) • -16/2 = -8 • (-8)2 = 64, • y = -5(x2 – 16x + 64) - 319 + 320 (WHY +320??) • y = -5(x – 8)2 + 1 • Vertex is (8, 1), Axis of Sym: x = 8, a < 0, so parabola turns downward

  17. y (3, 16) 4 (–1, 0) (7, 0) x 4 x = 3 Example: Graph and find the vertex and x-intercepts of f(x) = –x2 +6x + 7. Vertex and x-Intercepts f(x) = –x2 + 6x + 7 original equation f(x) = –(x2 – 6x) + 7 factor out –1 f(x) = –(x2 – 6x + 9) + 7 + 9 complete the square f(x) = –(x – 3)2 + 16 standard form a< 0parabola opens downward. h = 3,k = 16axis x = 3, vertex (3, 16). Find the x-intercepts by solving–x2 + 6x + 7 = 0. (–x + 7 )(x + 1) = 0 factor x = 7, x = –1 x-intercepts (7, 0), (–1, 0) f(x) = –x2 +6x + 7

  18. Vertex (3, 1) and Point (6, 3)

  19. Homework • WS 3-2

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