Can someone please help me with this question for a medal?????

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Can someone please help me with this question for a medal?????

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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1. Graph y = –2x2 + 6. Describe what you see. The graph looks like a parabola that is shifted up by around 6 units, and is stretched by a factor of 1/2. It is also pointing down, since the coefficient of -2x^2 is a negative. This question is complete, the next question is what I need help with graph y = (-1/10)x2 + 6. Describe what you see. How is what you observe here different from what you observed in part A?
1356
please add a medal!!!!

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1356?
Well, it is exactly the same way as you did the first time, but some things will change.
So is this graphed shifted by 5 instead of 6? Or is it also 6?
Im not sure how to graph that
Well, whenever you graph a second degree function, and it represents a parabola, it'll look like this: \[y=Ax^2 + Bx + C\] This is pretty much the equation of the parabola and we can find the orientation, or the cancavity if you like, by looking at the sign that "A" has, if it is negative, the parabola opens downwards, if "A" is positive, it opens upwards. So, in the function you were given: \[y=-\frac{ 1 }{ 10 }x^2+6\] "A" which in this case is "-1/10" is negative, so it opens downwards as well.
Oh okay. Makes sense. Alright, give me a second.
So when you say it opens downwards, you mean that it would look like a hill then?
yes, but there are more things you have to determine, like where it cuts the x-axis and where it cuts the y-axis.
Alright, I think I can handle it from here. Thank you!
No problem.

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