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

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More answers

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