Simple minimum/maximum question

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The discussion centers on finding the minimum point of the quadratic function y = x^2 + 10x + 19, which can be rewritten as y = (x + 5)^2 - 6. The minimum occurs at the vertex of the parabola, where the x-value is -5 (the opposite sign of +5) and the y-value is -6. The graph opens upwards, indicating there is no maximum point, as it continues to rise indefinitely. If the equation were to represent a downward-opening parabola, the maximum could be determined similarly. Understanding the vertex form of a quadratic is crucial for identifying minimum and maximum values.
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I did a past paper question whereby I carried the answer (X+5)-6 through.

it then asked for the minimum points of the graph y=x^2 + 10x + 19. which is what i made into (X+5)-6.

I know i have to take the +5 and change its sign to -. and that's the minimum of X. and the -6 without change is the minimum of Y.


I just wondered, can anyone explain to me why this is so?


Furthermore, what would I do if it asked for a maximum. I would show an attempt, but i don't even understand why this is a minimum so it's hard for me to find out the maximum.
( this isn't actually a question I'm just interested.)


thanks :P
 
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DeanBH said:
I did a past paper question whereby I carried the answer (X+5)-6 through.

it then asked for the minimum points of the graph y=x^2 + 10x + 19. which is what i made into (X+5)-6.
I presume you mean (x+5)2- 6.

I know i have to take the +5 and change its sign to -. and that's the minimum of X. and the -6 without change is the minimum of Y.


I just wondered, can anyone explain to me why this is so?


Furthermore, what would I do if it asked for a maximum. I would show an attempt, but i don't even understand why this is a minimum so it's hard for me to find out the maximum.
( this isn't actually a question I'm just interested.)


thanks :P
A square is never negative. If y= (x- 5)2- 6, It is always "-6 plus something". If x- 5= 0, which is the same as x= 5 (adding 5 to both sides. I cringe when I read somenthing like "take the+ 5 and change its sign to -"!), y= 0- 6= -6. For any other x, x- 5 is non-zero, (x- 6)2 is positive and (x- 5)2- 6 is larger that 6.

If the problem asked for a maximum, there is something wrong with the problem! The graph of y= (x- 5)2- 6 is a parabola that opens upward: its "vertex" is at the lowest point (5, -6). There is no highest point.

However, if the problem were y= -(x- 5)2- 6 (that's the same as y= x2+ 10 x+ 31), then you can argue that when x= 5, y= -02+ 31= 31 but for any other value of x, y= -(a positive number)+ 31 and so is less than 31. In this case, the graph is a parabola that opens downward. (5, 31) is the highest point on the parabola and 31 is the maximum value of y.
 
Can anyone tell me how i find the line of symmetry of this curve, and why it is that 8D
 
Do you know the definition of "line of symmetry"?
 
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Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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