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Graphing Quadratic Functions: Examples (page 3 of 4)

Sections: Introduction, The meaning of the leading coefficient / The vertex, Examples


  • Find the vertex and intercepts of y = 3x2 + x 2 and graph; remember to label the vertex and the axis of symmetry.

    This is the same quadratic as in the last example. I already found the vertex when I worked the problem above. This time, I also need to find the intercepts before I do my graph. To find the y-intercept, I set x equal to zero and solve:

      y = 3(0)2 + (0) 2
           = 0 + 0 2 = 2

    Then the y-intercept is the point (0, 2). To find the x-intercept, I set y equal to zero, and solve:

      0 = 3x2 + x 2
      0 = (3x 2)(x + 1)
      3x 2 = 0  or   x + 1 = 0
      x2/3   or  x = 1

    Then the x-intercepts are at the points (1, 0) and ( 2/3, 0).

    The axis of symmetry is halfway between the two x-intercepts at (1, 0) and at ( 2/3 , 0); using this, I can confirm the answer from the previous page:

      (1 + 2/3) / 2 = (1/3) / 2 = 1/6

    The complete answer is a listing of the vertex, the axis of symmetry, and all three intercepts, along with a nice neat graph:

       

    The vertex is at ( 1/6 , 25/12 ), the axis of symmetry is the line x = 1/6 , and the intercepts are at (0, 2), (1, 0), and ( 2/3, 0).

     

    graph of y = 3x^2 + x - 2 with vertex and intercepts marked.
     

  • Find the intercepts, the axis of symmetry, and vertex of y = x2 x 12.

    To find the y-intercept, I set x equal to 0 and solve:

      y = (0)2 (0) 12 = 0 0 12 = 12

    To find the x-intercept, I set y equal to 0 and solve:

      0 = x2 x 12

      0 = (x 4)(x + 3)

      x = 4  or  x = 3

    To find the vertex, I look at the coefficients: a = 1 and b = 1. Plugging into the formula, I get:

      h = (1)/2(1) = 1/2 = 0.5

    To find k, I plug h = 1/2 in for x inside y = x2 x 12, and simplify:

      k = (1/2)2 (1/2) 12 = 1/4 1/2 12 =  12.25

    Once I have the vertex, it's easy to write down the axis of symmetry: x = 0.5. Now I'll find some additional plot points, to fill in the graph:

      T-chart

    For my own convenience, I picked x-values that were centered around the x-coordinate of the vertex. Now I can plot the parabola:   Copyright Elizabeth Stapel 2002-2011 All Rights Reserved

      graph of y = x^2 - x - 12

    The vertex is at the point (0.5, 12.25),
       the axis of symmetry is the line
    x = 0.5,
          
    and the intercepts are at the points (0, 12)(3, 0), and (4, 0).

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Cite this article as:

Stapel, Elizabeth. "Graphing Quadratic Functions: Examples." Purplemath. Available from
    http://www.purplemath.com/modules/grphquad3.htm. Accessed
 

 



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