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Findind the vertex?having 2 different answer?

Topic: Different methods in problem solving
July 15, 2019 / By Harmony
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

Dominique | 1 day ago
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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We found more questions related to the topic: Different methods in problem solving

Originally Answered: If you have a vertex of (4, -14) whats the Minimum Value, I put 4?
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
Originally Answered: If you have a vertex of (4, -14) whats the Minimum Value, I put 4?
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.
Originally Answered: If you have a vertex of (4, -14) whats the Minimum Value, I put 4?
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

Carrie
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Originally Answered: How to find the vertex and focus?