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kylewendt

  • 2 years ago

How do you graph the quadratic equation 3x^2=3? I need help, my EOC is next Tuesday.

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  1. TurtleMuffin
    • 2 years ago
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    First you can divide both sides by 3 to get x^2 = 1. Subtract 1 from both sides and you have x^2 - 1 = 0. The roots of this are -1 and 1. So this is a standard parabola shifted down one united passing through -1 and 1 on the x-axis.

  2. FibonacciChick666
    • 2 years ago
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    first, what shape does the graph make?

  3. kylewendt
    • 2 years ago
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    The graph makes a parabola when completed

  4. FibonacciChick666
    • 2 years ago
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    good, now, that is an important concept, so now, you have one slight problem, This is only in one variable(x) so you can't graph it. Are you sure you copied the question correctly?

  5. kylewendt
    • 2 years ago
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    Yes. The exact question is: What are the solutions of the equation 3x^2=3? Use a graph of the related function.

  6. Hero
    • 2 years ago
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    @FibonacciChick666, what do you mean it can't be graphed?

  7. FibonacciChick666
    • 2 years ago
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    'graphed' I mean, it's a number line as stated

  8. Hero
    • 2 years ago
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    Continuing what Turtlemuffin was saying, you have x^2 - 1 = 0 which factors to (x + 1)(x - 1) = 0

  9. kylewendt
    • 2 years ago
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    My question is, after you have your two x-intercepts, how do you find the vertex from (x+1)(x-1)=0?

  10. Hero
    • 2 years ago
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    We can use the vertex formula to graph the vertex. \(x = -\dfrac{b}{2a}\)

  11. Hero
    • 2 years ago
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    In this case b = 0 and a = 1, so \[x = -\frac{0}{2(1)} = 0\]

  12. FibonacciChick666
    • 2 years ago
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    @Hero , one variable= 1-D , no?

  13. Hero
    • 2 years ago
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    Since \(x = 0\), we can insert 0 into the expression to get y = \(0^2 - 1 = -1\) so the vertex is \((0,-1)\)

  14. kylewendt
    • 2 years ago
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    So, the graph would look like this?

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  15. TurtleMuffin
    • 2 years ago
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    That is an algebraic approach. You should know the graph of y=x^2. It's the standard parabola. We simplified your given equation into 0 = x^2 - 1. The "minus 1" tells us our standard parabola shifts down one unit. Therefore its x-coordinate remains the same, and it's y-coordinate is shifted down 1 unit from 0, i.e. it's now -1. You can remember all the formulas you want, but if you don't see how an algebraic equation is related to its graphical representation, you won't enjoy math too much. That is has how the graph looks like though!

  16. FibonacciChick666
    • 2 years ago
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    ah, now that I re-read the prompt, I understand.

  17. FibonacciChick666
    • 2 years ago
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    It wanted the roots, by looking at the related graph

  18. kylewendt
    • 2 years ago
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    Thank you all for explaining, as I understand this concept now. My algebra teacher is one of the worst (I know that sounds generic), and intentionally ignores my questions to her on assignments because she doesn't like me. This really helps.

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