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April 27, 2026, 06:47:13 pm
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Author Topic: Help again?  (Read 926 times)  Share 

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VVVCCCEEE

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Help again?
« on: March 13, 2018, 02:57:22 pm »
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Idk if I'm stupid or what but how question b get (-1/2, -1/4) as the tp?
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lazaward

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Re: Help again?
« Reply #1 on: March 13, 2018, 03:54:09 pm »
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1. Complete the square for the function.
f(x) = (x^2 + x)
f(x) = (x^2 + x + (1/2)^2) - (1/2)^2))
f(x) = (x + 1/2)^2 - (1/2)^2
f(x) = (x + 1/2)^2 - 1/4
2. Now that it’s in TP form we can see that the TP is (-1/2, -1/4).
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VVVCCCEEE

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Re: Help again?
« Reply #2 on: March 13, 2018, 04:34:32 pm »
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Thank youu
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TheAspiringDoc

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Re: Help again?
« Reply #3 on: March 13, 2018, 07:38:07 pm »
+1
Or
x2+x
=x(x+1)
So x=-1 or x=0 at x-ints
TPx = -b/(2a) for ax2+bx+c (I.e. Is halfway between the x-ints)
TPx = -1/2
Sub that into the original function to get the y-int.

Carolinesale's method is perfectly fine, I just think that this way might be faster as you can do it in your head
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