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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?
 one year ago
 one year 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?
 one year ago
 one year ago

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Comm.DanBest ResponseYou've already chosen the best response.0
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.
 one year ago

Comm.DanBest ResponseYou've already chosen the best response.0
Here is the problem y equals x squared + 1/2x + 2
 one year ago

Comm.DanBest ResponseYou've already chosen the best response.0
The Quadratic Formula is x equals b + or  square root sign then b squared 4ac over 2a
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
the vertex is along the x = b/(2a) axis therefore the vertex itself is simply (b/(2a), f(b/(2a)))
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
\[y = (x)^2 + \frac{1}{2x} + 2\] what is your equation?
 one year ago

Comm.DanBest ResponseYou've already chosen the best response.0
The equation is y equals x squared + 1/2x + 2
 one year ago

harsimran_hs4Best ResponseYou've already chosen the best response.0
are you trough with the question or still some confusion?
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
youre restating the equation in the same manner that you first presented it does not clarify it in any way whatsoever ...
 one year ago

Comm.DanBest ResponseYou've already chosen the best response.0
I get how to get the first answer but not the second answer @amistre64 and @harsimran_hs4
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
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. \[\left(\frac{\sqrt{b^24ac}}{2a}\right)\leftarrow\leftarrow \left(\frac{b}{2a}\right)\to \to\left(\frac{+\sqrt{b^24ac}}{2a}\right)\] 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 \(b^24ac\) is a positive value; we have 2 equal distances from the center that define the x intercepts. If \(b^24ac\) is equal to 0, then we only have one point on the x axis, the vertex itself. If \(b^24ac\) 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.
 one year ago
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