Finding the directrix, focus, and roots of a parabola Help!  TOPIC_SOLVED

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Finding the directrix, focus, and roots of a parabola Help!

Postby nick5892 on Fri Apr 10, 2009 6:29 pm

Find the directrix, focus , and roots of the parabola y= x^2 - 5x + 4....... ok here is where I am so far

First thing i did was isolate the x variables so i can complete the square.

y - 4= x^2 - 5x

Then i completed the square...

y-4 = x^2 - 5x half of - 5= -5/2 and squaring that i got 25/4 so i added that to both sides....

y-4 + 25/4 = x^2 -5x + 25/4

simplified....

y+9/4 = ( x- 5/2) ^2


and now i dont know how to get my "a" to complete my parabola equation y-k = a(x-h)^2 .....i can solve the problem if i knew how to get it... :confused: please help!
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Postby stapel_eliz on Fri Apr 10, 2009 8:49 pm

First, thank you for showing your work so nicely! :thumb:

nick5892 wrote:Find the directrix, focus , and roots of the parabola y= x^2 - 5x + 4....... ok here is where I am so far First thing i did was....

...y+9/4 = ( x- 5/2) ^2

and now i dont know how to get my "a" to complete my parabola equation y-k = a(x-h)^2 .

The "a" in "y - k = a(x - h)^2" is the same "a" as in "y = ax^2 + bx + c". Thus, in your case, a = 1. :wink:
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Re: Finding the directrix, focus, and roots of a parabola Help!

Postby reubenator on Sun Apr 12, 2009 2:52 pm

I just learned this but for future help the "a" is always in front of the squared term.
so if its 2(x-3)^2 your "a" is 2.
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Re: Finding the directrix, focus, and roots of a parabola Help!

Postby nick5892 on Mon Apr 13, 2009 12:13 am

thanks for your help guys! kind of obvious now that i see the answer :lol:
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