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Topic: **Different methods in problem solving****Question:**
Im a bit confuse on this vertext problem
Y= -(-x-6)(x+2)
I know how to solve this,
But my answer comes out different.
The first answer came out to be 2
-(x^2+2x-6x-12)
-x^2-2x+6x+12
-X^2+4x+12
Which is 2
But whe i plug 2 back in to the equation
-X^2+4x+12
I got 24
-(2)^2+4(2)+12
However the answer is 16, they plug the 2 into
-(-x-6)(x+2) but isnt the answer suppose to come out the same?
Thanks

July 15, 2019 / By Harmony

you did the wrong thing. as per your solution you must do it as... -[-x^2-2x-6x-12] x^2+8x+12 right one another method cancel out the negative sign first by taking -1 common and than multiply . and vertex will be (-4,-4) to find vertex you can use calculus too.. i.e maximization or minimisation first diffrentiate it and put it =0 find value of x you dont need to check that it is maximum or minimum.. just put the vale in the above and you get value of y. diffr. 2x+8=0 x=-4 putting in above y = -4

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Hi, No, sorry, you were wrong. The y-value, -14, is the minimum value of the function. The x-value, 4, just tells where the vertex (or any other y-value) falls on the x-axis. Incidentally, in order for the function to open downward, the x-term would need to be negative. In that case, the function would have a maximum value, but no minimum unless there was a specified domain. Hope this helps. FE

I don't quite understand your question. but if it's the first formula you wrote the vertex is (3, Something) making the max. 3. if it's the second one it makes the vertex (4, -14) making the max value 4. On a down-facing parabola it's not possible to have a minimum value. It must be a maximum.

i have already answered the other "general " question...if you have y = a [x-h]² + k and a > 0 then minimum value of y is k , taken when x = h...so 4 is not correct , nor is 3 for the 1st remembrance

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I just have to respond to this Q because as I now look at the answer that MisterGreen gave, I think he messed it up. How he moved the squared term to be the "x" term is a mystery to me. This makes a big difference. You started off solving this problem correctly. You completed the square correctly to get it down to: (y+3)^2 = x - 7 But then you went off the rails a little bit. It looks like you tried to "complete the square" again, but you don't have to do that at this point. How you proceed at this point depends on how your textbook handles conics. Some books "solve" the equation for the variable that's not squared, and some just leave it like I have it above. I guess I'll try to do it both ways and you can just look at the one that matches up with how your book works out these problems. Method 1 (= solving for the un-squared variable) - You can just move the -7 over to the left side to get: (y+3)^2 + 7 = x And now you can just look at what you have and think about it some to learn some info. about this parabola. Since the "y" term is getting squared, you've got the formula of a parabola that's lying on its side. And although it doesn't look like it, there's a positive 1 in front of the (y+3)^2 term. This tells you that the parabola opens up to the right since its positive. The general formula used here is, x = a(y – k)^2 + h, where the vertex is at (h, k). If you look at this formula, there's a negative sign in the term that's squared. Yours doesn't have that yet, so you can re-write it this way, x = [y - (-3)]^2 + 7 And now, carefully comparing this with the general formula, you can see that the vertex is at (7,-3). And then to find the focus, you can use the following formula, since in this situation the focus lies directly to the right of the vertex: ( h + 1/4a, k). For yours, "a" equals 1, so the focus is at (7 + 1/4, -3), or you can write this as (7 1/4, -3) or in decimal form it would be (7.25, -3). Method 2 (= writing the equation with just the squared term on one side) - With this method, books like to use the letter "p" instead of "a", and they write the general equation as: 4p(x – h) = (y – k)^2. With this one, there are negative signs in both places. Writing yours in this format gives, x - 7 = [y - (-3)]^2. And again the "formula" for the vertex is (h, k), so your vertex is at (7, -3). And now for the focus, it's a little different than above. With this method, the books say that the formula for the focus of a parabola lying on its side is, (h + p, k). To find "p" here, it might be helpful to write your equation like this: 1(x - 7) = [y - (-3)]^2. You can now see better that 4p = 1. And solving this for "p" gives: p = 1/4. Using this in (h + p, k), gives the same focus as above, (7 1/4, -3). I hope this helped to get you un-lost. Parabolas can be tricky because they can open up (or down), or they can be on their sides. Trying to picture what you have in your head, would probably help keep things straight. And remembering the general formats is also needed. Jack (4-16-14)

x^2 - 4x - 16 = 6y <- Moved the Y x^2 - 4x = 6y + 16 <- Moved the number (x - 2)^2 = 6y + 20 <- Simplified (x - 2)^2 - 20 = 6y (1/6)(x - 2)^2 - 10/3 = y vertex: (2, -10/3) <- Answer

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