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6.6 analyzing graphs of quadratic functions
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6.6 analyzing graphs of quadratic functions

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  • 1. 6.6 Analyzing Graphs of Quadratic Functions
  • 2. Graphing the parabola y = f ( x ) = ax 2
    • Consider the equation y = x 2
    0 1 4 1 (–1, 1) (0, 0) (1, 1) (2, 4) y x 4 (–2, 4) Axis of symmetry: x = 0 ( y=x 2 is symmetric with respect to the y-axis ) Vertex (0, 0) When a > 0 , the parabola opens upwards and is called concave up. The vertex is called a minimum point. y 2 1 0 – 1 – 2 x
  • 3.
    • For the function equation y = x 2 , what is a ?
    a = 1 . What if a does not equal 1? Consider the equation y = – 4 x 2 . 0 – 4 – 4 (0, 0) – 16 y – 16 (–2, –16) (–1, –4) (1, –4) (–4, –16) x When a < 0 , the parabola opens downward and is called concave down. The vertex is a maximum point. What is a ? a = – 4 y 2 1 0 – 1 – 2 x
  • 4. Properties of the Parabola f ( x ) = ax 2
    • The graph of f ( x ) = ax 2 is a parabola with the vertex at the origin and the y axis as the line of symmetry.
    • If a is positive , the parabola opens upward , if a is negative , the parabola opens downward .
    • If  a  is greater than 1 (  a  > 1), the parabola is narrower then the parabola f ( x ) = x 2 .
    • If  a  is between 0 and 1 (0 <  a  < 1), the parabola is wider than the parabola f ( x ) = x 2 .
  • 5. Graphing the parabola y = f ( x ) = ax 2 + k Algebraic Approach: y = – 4 x 2 – 3 Numerical Approach: Graphical Approach: Consider the equation y = – 4 x 2 – 3 . What is a ? a = – 4 x Vertex (0, -3) -16 -4 0 -4 -16 -4 x 2 -19 -7 -3 -7 -19 - 4 x 2 - 3 2 1 0 – 1 – 2 x
  • 6. y = – 4 x 2 – 3 . x Vertex (0, -3) y = – 4 x 2 In general the graph of y = ax 2 + k is the graph of y = ax 2 shifted vertically k units. If k > 0, the graph is shifted up. If k < 0, the graph is shifted down. (P. 267) The graph y = – 4 x 2 is shifted down 3 units.
  • 7. a = – 4. What effect does the 3 have on the function? y x y = – 4 x 2 y = – 4( x – 3) 2 Consider the equation y = – 4( x – 3) 2 . What is a ? The axis of symmetry is x = 3. Numerical Approach: Axis of symmetry is shifted 3 units to the right and becomes x = 3 -16 -4 0 -4 -16 – 4 x 2 2 1 0 – 1 – 2 x -4 -16 -36 -64 -100 - 4 ( x- 3) 2 -36 3 0
  • 8. Vertex Form of a Quadratic Function
    • The quadratic function
    • f ( x ) = a ( x – h ) 2 + k, a 0
    • The graph of f is a parabola .
    • Axis is the vertical line x = h.
    • Vertex is the point ( h, k ).
    • If a > 0, the parabola opens upward.
    • If a < 0, the parabola opens downward.
  • 9. 6.6 Analyzing Graphs of Quadratic Functions
    • A family of graphs – a group of graphs that displays one or more similar characteristics.
      • Parent graph – y = x 2
    • Notice that adding a constant to x 2 moves the graph up .
    • Notice that subtracting a constant from x before squaring it moves the graph to the right .
    y = x 2 y = x 2 + 2 y = (x – 3) 2
  • 10. Vertex Form
    • Each function we just looked at can be written in the form (x – h) 2 + k, where (h , k) is the vertex of the parabola, and x = h is its axis of symmetry.
      • (x – h) 2 + k – vertex form
    x = 3 ( 3 , 0 ) y = (x – 3 ) 2 or y = (x – 3 ) 2 + 0 x = 0 ( 0 , 2 ) y = x 2 + 2 or y = (x – 0 ) 2 + 2 x = 0 ( 0 , 0 ) y = x 2 or y = (x – 0 ) 2 + 0 Axis of Symmetry Vertex Equation
  • 11. Graph Transformations
    • As the values of h and k change, the graph of y = a(x – h) 2 + k is the graph of y = x 2 translated.
      • | h | units left if h is negative, or |h| units right if h is positive.
      • | k | units up if k is positive, or | k | units down if k is negative.
  • 12. Graph a Quadratic Function in Vertex Form
    • Analyze y = (x + 2) 2 + 1. Then draw its graph.
    • This function can be rewritten as y = [x – (-2)] 2 + 1.
      • Then, h = -2 and k = 1
    • The vertex is at (h , k) = (-2 , 1), the axis of symmetry is x = -2. The graph is the same shape as the graph of y = x 2 , but is translate 2 units left and 1 unit up.
    • Now use the information to draw the graph.
    • Step 1 Plot the vertex (-2 , 1)
    • Step 2 Draw the axis of symmetry, x = -2.
    • Step 3 Find and plot two points on one side of the axis symmetry, such as (-1, 2) and (0 , 5).
    • Step 4 Use symmetry to complete the graph.
  • 13. Graph Transformations
    • How does the value of a in the general form y = a(x – h) 2 + k affect a parabola? Compare the graphs of the following functions to the parent function, y = x 2 .
    • a.
    • b.
    • c.
    • d.
    y = x 2 a b d c
  • 14. Example: Write a quadratic function for a parabola with a vertex of (-2,1) that passes through the point (1,-1).
    • Since you know the vertex, use vertex form! y=a(x-h) 2 +k
    • Plug the vertex in for (h,k) and the other point in for (x,y). Then, solve for a.
    • -1=a(1-(-2)) 2 +1
    • -1=a(3) 2 +1
    • -2=9a
    Now plug in a, h, & k!
  • 15. Standard 9 Write a quadratic function in vertex form Vertex form- Is a way of writing a quadratic equation that facilitates finding the vertex. y – k = a ( x – h ) 2
    • The h and the k represent the coordinates of the vertex in the form V( h, k).
    • The “a” if it is positive it will mean that our parabola opens upward and if negative it will open downward.
    • A small value for a will mean that our parabola is wider and vice versa.
  • 16. Standard 9 Write a quadratic function in vertex form Write y = x 2 – 10 x + 22 in vertex form. Then identify the vertex. y = x 2 – 10 x + 22 Write original function. y + ? = ( x 2 – 10 x + ? ) + 22 Prepare to complete the square. y + 25 = ( x 2 – 10 x + 25 ) + 22 y + 25 = ( x – 5) 2 + 22 Write x 2 – 10 x + 25 as a binomial squared. y + 3 = ( x – 5) 2 Write in vertex form. Add – 10 2 2 ( ) = (–5) 2 = 25 to each side. The vertex form of the function is y + 3 = ( x – 5) 2 . The vertex is (5, – 3). ANSWER
  • 17. EXAMPLE 1 Write a quadratic function in vertex form Write a quadratic function for the parabola shown. SOLUTION Use vertex form because the vertex is given. y – k = a ( x – h ) 2 Vertex form y = a ( x – 1) 2 – 2 Substitute 1 for h and –2 for k . Use the other given point, ( 3 , 2 ) , to find a . 2 = a ( 3 – 1) 2 – 2 Substitute 3 for x and 2 for y . 2 = 4 a – 2 Simplify coefficient of a . 1 = a Solve for a .
  • 18. EXAMPLE 1 Write a quadratic function in vertex form A quadratic function for the parabola is y = ( x – 1) 2 – 2 . ANSWER
  • 19. EXAMPLE 1 Graph a quadratic function in vertex form Graph y – 5 = – ( x + 2) 2 . SOLUTION STEP 1 STEP 2 Plot the vertex ( h , k ) = ( – 2, 5) and draw the axis of symmetry x = – 2. 14 Identify the constants a = – , h = – 2, and k = 5. Because a < 0, the parabola opens down. 14
  • 20. EXAMPLE 1 Graph a quadratic function in vertex form STEP 3 Evaluate the function for two values of x . Plot the points (0, 4) and (2, 1) and their reflections in the axis of symmetry. STEP 4 Draw a parabola through the plotted points. x = 0 : y = ( 0 + 2) 2 + 5 = 4 14 – x = 2 : y = ( 2 + 2) 2 + 5 = 1 14 –
  • 21. GUIDED PRACTICE for Examples 1 and 2 Graph the function. Label the vertex and axis of symmetry. 1. y = ( x + 2) 2 – 3 2. y = –( x + 1) 2 + 5
  • 22. GUIDED PRACTICE for Examples 1 and 2 3. f ( x ) = ( x – 3) 2 – 4 12