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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?

Mathematics
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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.
Here is the problem y equals -x squared + 1/2x + 2
The Quadratic Formula is x equals -b + or - square root sign then b squared -4ac over 2a

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Other answers:

the vertex is along the x = -b/(2a) axis therefore the vertex itself is simply (-b/(2a), f(-b/(2a)))
\[y = (-x)^2 + \frac{1}{2x} + 2\] what is your equation?
The equation is y equals -x squared + 1/2x + 2
are you trough with the question or still some confusion?
youre restating the equation in the same manner that you first presented it does not clarify it in any way whatsoever ...
I get how to get the first answer but not the second answer @amistre64 and @harsimran_hs4
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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