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Pre-cal graph

  1. Feb 1, 2004 #1
    I'm missing a step somewhere.

    Without a calculator, graph y=3x^2-16x-12 by factoring and plotting zeros.

    I have gotten as far as (x=-2/3) & (x=6) and know it's a parabola and pointing up because it has the positive x^2 so the graphing is easy enough, except...

    The botb says the y-int is -100/3 but I can't find the bridge to get there in my notes or in the chapter.

    a hint, please.

    Thank you.
  2. jcsd
  3. Feb 1, 2004 #2

    Doc Al

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    Staff: Mentor

    Must be a misprint. The y-intercept is the point where the curve intersects the y-axis, in other words: the value of y where x=0. That value is certainly not y = -100/3.
  4. Feb 2, 2004 #3
    Sorry, my error. The bottom coordinate of the parabola is
    (8/3,-100/3), not the y-int.

    Thank you for your prompt reply.

  5. Feb 2, 2004 #4


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    Staff Emeritus
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    In other words, the vertex is at (8/3,-100/3).

    You can find that by completing the square.
    y=3x2-16x- 12= 3(x2- (16/3)x)- 12.
    (16/3)/2= 8/3 and (8/3)2= 64/9
    y= 3(x2- (16/3)x+ 64/9- 64/9)- 12
    = 3(x2- (16/3)x+ 64/9)- 64/3- 12
    = 3(x-8/3)2- 100/3

    Now it is clear that when x= 8/3, y= -100/3. And that if x is any other number, then y= -100/3 plus something and so is higher. (8/3, -100/3) is the lowest point on the graph- the vertex.
  6. Feb 2, 2004 #5
    I'm vaaaaaguely familiar with completing the square. It was briefly touched on in my trig class.

    I understand that if x is any other number <,> 8/3 that it will follow the parabola and make the y value change to a number greater than -100/3. Thanks.

    This afternoon, one of the schools free tutors showed me to use -b/2a. Where does this little gem come from?
  7. Feb 3, 2004 #6


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    It's for those who prefer memorizing formulas rather than thinking!

    If y= ax2+ bx+ c, then y= a(x2+ (b/a)x)+ c.

    Now, think "(b/a)/2= b/(2a) and that squared is b2/4a2" so we need to add b2/4a2 to complete the square:

    y= a(x2+ (b/a)x+ b2/4a2-b2/4a2
    = a(x2+ (b/a)x+ b2/4a2)- b2/4a+ c
    = a(x+ b/(2a))2+ (c- b2/4a)

    Now, we can see that, when x= -b/(2a), y= c- b2/4a. If x is any other number, the square is positive so y is larger. The vertex of the parabola (the lowest point) is at (-b/2a,c- b2/4a).

    I think that completing the square is important enough that you should know how to do it, and practice it, with out just memorizing that formula for the vertex.
  8. Feb 3, 2004 #7
    Thanks Halls. Exam today, this should come in handy.

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