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Comm.Dan

  • 3 years ago

How would you find the vertex of this quadratic function either by finding the vertex using a, b, and c or finding the vertex by using just a and b?

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  1. Comm.Dan
    • 3 years ago
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    To find using a, b, and c, you use the Quadratic Formula. To find using a and b you use x equals negative b over 2a, then you put the x back into the original quadratic function to find the y.

  2. Comm.Dan
    • 3 years ago
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    Here is the problem y equals -x squared + 1/2x + 2

  3. Comm.Dan
    • 3 years ago
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    The Quadratic Formula is x equals -b + or - square root sign then b squared -4ac over 2a

  4. amistre64
    • 3 years ago
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    the vertex is along the x = -b/(2a) axis therefore the vertex itself is simply (-b/(2a), f(-b/(2a)))

  5. amistre64
    • 3 years ago
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    \[y = (-x)^2 + \frac{1}{2x} + 2\] what is your equation?

  6. Comm.Dan
    • 3 years ago
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    The equation is y equals -x squared + 1/2x + 2

  7. Comm.Dan
    • 3 years ago
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    @amistre64

  8. harsimran_hs4
    • 3 years ago
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    are you trough with the question or still some confusion?

  9. amistre64
    • 3 years ago
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    youre restating the equation in the same manner that you first presented it does not clarify it in any way whatsoever ...

  10. Comm.Dan
    • 3 years ago
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    I get how to get the first answer but not the second answer @amistre64 and @harsimran_hs4

  11. amistre64
    • 3 years ago
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    the structure of the quadratic formula informs us about the axis of symmetry (the middle of the parabola) as well as the x intercepts (which are equal distances to the left and right of the axis of symmetry). All these parts reflect the value of the x components of the parabola. \[\left(\frac{-\sqrt{b^2-4ac}}{2a}\right)\leftarrow\leftarrow \left(\frac{-b}{2a}\right)\to \to\left(\frac{+\sqrt{b^2-4ac}}{2a}\right)\] Since the vertex of the parabola lies on the axis of symmetry, the value of the x component of the vertex is: \(\Large \frac{-b}{2a}\) . Can you see what happens to the left and right parts of this? If \(b^2-4ac\) is a positive value; we have 2 equal distances from the center that define the x intercepts. If \(b^2-4ac\) is equal to 0, then we only have one point on the x axis, the vertex itself. If \(b^2-4ac\) is a negative value, then there is no points of the parabola touching the x axis and therefore it is floating someplace above or below it.

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