equation for the translated graph below (Pre-Calc)
- Thread starter reddec85
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Harry_the_cat
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Have another think about the points you have as dots on the first graph. They look like they are at (1, 4) and (-1, 4). Is that correct?
Dr.Peterson
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Just wondering where I got stuck here, looks like my method was correct until the end. I’d ask my teacher but Holiday Break and all that
Pay close attention to the order of operations. The function is [imath]f(x) = 2x^2[/imath], not [imath]f(x) = (2x)^2[/imath]. Do you see the difference?
Ah, I see what I did. I’ve actually made a mistake like this in the past, where I’ve forgotten to do the exponent first. Thanks for your help!
Harry_the_cat
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Besides having the incorrect points, your method is correct.
However, there is a muck quicker way:
Reflecting \(\displaystyle y=2x^2\) in the x-axis would give \(\displaystyle y=-2x^2\).
Then moving the graph \(\displaystyle y=-2x^2\) up 5 units would give \(\displaystyle y=-2x^2 + 5\).
{Note: There actually is a mistake in your method. The +k should lead to +5 not -5.}
However, there is a muck quicker way:
Reflecting \(\displaystyle y=2x^2\) in the x-axis would give \(\displaystyle y=-2x^2\).
Then moving the graph \(\displaystyle y=-2x^2\) up 5 units would give \(\displaystyle y=-2x^2 + 5\).
{Note: There actually is a mistake in your method. The +k should lead to +5 not -5.}