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Mth163
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symmetry and parabolas
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Question: `q007. Sketch a parabola through points (-4, 4), (1, -1) and (2, 4). The parabola should be symmetric about some vertical line.
Estimate the coordinates at which the parabola passes through the x axis.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
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If the parabola is at (1,-1) then the line of symmetry should pass through x=1
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There is no reason to expect that the vertex is at (1, -1).
In fact the other two point dictate that the vertex lies elsewhere.
When you sketched the parabola where was your vertex? If it was at (1, -1) then your parabola would not have been symmetric.
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I thought all parabolas were always symmetric
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They are, but that doesn't mean the the vertex was among the coordinates I gave you.
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How far do you have to go to the right, starting at (1, -1), to get to (2, 4)?
How far do you have to go to the left to get to (-4, 4)?
How far is it between (-4, 4) and (2, 4)?
How far should you have to go to the right from (-4, 4) to get to the vertex?
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#$&*
Mth163
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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y=at^2+bt+c do we square the a
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y=at^2+bt+c do we square the a as well of just the t value?
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As dictated by the order of operations, you don't square the a.
Be sure you understand why, in terms of order of operations.
Related Note: For an expression like -5^2, you square the 5 but not the - sign. -5^2 = -25. If you want to square -5, you write it (-5)^2.
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#$&*
Mth163
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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video clips
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The video clips have not worked since after depth vs time class notes, is that a problem or is it supposed to be like that (as in there are no more class videos)???
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As far as I know everything works.
We can check this out at lunchtime.
Remind me.
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