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## #1 2020-08-24 08:32:59

Otsdarva
Member
Registered: 2010-12-21
Posts: 10

Faced with this quadratic I was tasked with finding the vertex and y-intercept: -3/2x^2-6x

I could just use the formula -b/2a but because I'm learning to fully understand rather than just memorize shortcuts, I take the long way.  Get rid of -3/2 by multiplying everything by -2/3 and left with: x^2+4x

I complete the square to get (x+2)^2-4 (half of 4 is 2, 2 squared is 4, subtract a 4 to maintain equality).. but this is wrong and I still don't know why it's wrong.  This gives me the vertex (-2,-4) but the correct vertex is (-2,6).  Where am I messing up?  The only way I can reach (-2,6) as an answer is if I had -2/3 on the other side of the equation in the first step, to end up with x^2+4x=-2/3.  Then multiply the lone -4 at the end by -3/2.. but isn't that not possible?  If the equation is set to equal zero and I multiply everything by -2/3 in the beginning, wouldn't I just be left with zero on the other side of the equation?

Last edited by Otsdarva (2020-08-24 08:34:55)

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## #2 2020-08-24 19:23:01

Bob
Registered: 2010-06-20
Posts: 8,914

hi Otsdarva

My method is

From this you get roots of -4 and 0, and hence the x coordinate of the vertex is -2, leading to a y coordinate of +6.

Taking out the -3/2 is OK but you need to maintain the factor as you complete the square, otherwise you are changing the quadratic.

Bob

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

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## #3 2020-08-25 04:15:16

Otsdarva
Member
Registered: 2010-12-21
Posts: 10