Forget No More: How to Find a Turning Point in a Quadratic Graph

AI Thread Summary
The turning point of a quadratic graph, known as the vertex, can be found using the formula x = -b/(2a), applicable to the standard form y = ax² + bx + c. Additionally, the vertex can be determined by completing the square or using second-order differentiation. Understanding these methods is crucial for accurately identifying the turning point in quadratic equations. Mastering these techniques enhances one's ability to analyze and graph quadratic functions effectively. Finding the turning point is essential for solving various mathematical problems involving parabolas.
Michael17
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I seem to have forgotten the rule to find a Turning point in a quadratic graph. Can anyone help me?
 
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The turning point on a quadratic graph is at the vertex, whose x-coordinate is given by x=\frac{-b}{2a}
 
That's assuming that the quadratic is y= ax2[/sub]+ bx+ c!

You can find the turning point (vertex) by completing the square.
 
Or by 2nd order differentiation.

The Bob
 

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I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
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Thread 'Greatest possible value of a constant in polynomial'

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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