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anonymous

  • 5 years ago

graph the parabola. showt the zeros and the maxima or minima. y=x^2-6x+8

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  1. anonymous
    • 5 years ago
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    you find the zeros when you solve x^2 - 6x + 8 = 0 you need the first derivative for maxima and minima the same thing... first derivative = 0

  2. anonymous
    • 5 years ago
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    that is true if this is a calculus question. However the question may be answered without calculus The max or min of a parabola is the vertex, the axis of symmetry is given by \[x=\frac{-b}{2a}\] for the parabola in the form \[y=ax^2+bx+c\] get the vertex by evaluating for y at this x value moreover this parabola will have a min as the coefficient of x^2 is positive

  3. anonymous
    • 5 years ago
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    A plot of the parabola is attached.

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  4. anonymous
    • 5 years ago
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    thanks yalll

  5. anonymous
    • 5 years ago
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    robtobey howd you get that?

  6. anonymous
    • 5 years ago
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    i have to show work :/

  7. anonymous
    • 5 years ago
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    The plot was generated by Mathematica 8 with this statement: Plot[x^2-6x+8,{x,1,5}] You might Google "how to plot a parabola" I have never plotted a curve without Mathematica for years.

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