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Re: If y = x2 + ax + b, y is minimum when x is:
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27 Dec 2015, 06:16
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2
shreyashid wrote:
If y = x2 + ax + b, y is minimum when x is:
a) a/b b) -a/b c) -a/2 d) -b/2 e) b/a
I tried it by substituting the value of x everywhere For once that is making the problem lengthy, secondly I got stuck . Can anybody help please?
Hi, the easiest way is algebrically... if your quad eq is a*x^2 + b*x +c, its min or max value will occur at -b/2a... since the eq forms a parabola/ curve due to value a*x^2, the max or min value will depend on a.. here the eq is x^2 + ax + c=y.. so b in -b/2a...= a in x^2 + ax + c and a=1.. so -b/2a will become -a/2 for this equation .. ans C..
_________________
Re: If y = x2 + ax + b, y is minimum when x is:
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29 Dec 2015, 03:08
shreyashid wrote:
If y = x^2 + ax + b, y is minimum when x is:
a) a/b b) -a/b c) -a/2 d) -b/2 e) b/a
I tried it by substituting the value of x everywhere For once that is making the problem lengthy, secondly I got stuck . Can anybody help please?
Theory: Assume the equation to be A\(x^2\) + Bx + C = 0 The minimum value of this equation is found by differentiating it once and putting = 0 Hence 2Ax + B = 0 Therefore the minimum value of this equation will be at x = \(\frac{-B}{{2A}}\)
Coming back to the problem, if we compare x^2 + ax + b with A\(x^2\) + Bx + C = 0 we have A = 1, B = a and C = b
Therefore the minimum value will be at x = -a/2*1 = -a/2
Re: If y = x2 + ax + b, y is minimum when x is:
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15 Sep 2016, 03:16
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The minimum value of this equation is found by differentiating it once and putting = 0 Hence 2Ax + B = 0 how did you reach to 2Ax+B=0.Pls explain in detail.srt for the troube
Re: If y = x2 + ax + b, y is minimum when x is:
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18 Feb 2017, 05:52
shreyashid wrote:
If y = x^2 + ax + b, y is minimum when x is:
a) a/b b) -a/b c) -a/2 d) -b/2 e) b/a
I tried it by substituting the value of x everywhere For once that is making the problem lengthy, secondly I got stuck . Can anybody help please?
Hi,
We can use the method of completing the square to solve this problem. \(\begin{align*} y &= x^{2} + ax +b\\ &= x^{2} + 2\times x \times \frac{a}{2} + \left(\frac{a}{2}\right)^{2} + b - \left(\frac{a}{2}\right)^{2}\\ &= \left( x + \frac{a}{2} \right)^{2} + b - \left(\frac{a}{2}\right)^{2} \end{align*}\)
The above expression will have minimum value at \(x = -\frac{a}{2}\).
Re: If y = x2 + ax + b, y is minimum when x is:
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31 Mar 2019, 20:34
Please remember the basic rule for quadratic equations: ax^2 + bx + c = y Here since y is an upward facing parabola if a>0 we'll have minimum value -b/2a (max value reaching +ve infinity), similarly if a<0 we'll have a downward facing parabola with maximum value -b/2a (min value reaches -ve infinity).
Hence the answer will be -b/2a i.e., -a/2
gmatclubot
Re: If y = x2 + ax + b, y is minimum when x is:
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31 Mar 2019, 20:34
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58% (01:42) correct 
