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the minimum value

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by sanju09 » Sat Apr 04, 2009 3:14 am
What is the minimum value of the expression x ^ 2 + 6 x + 4?
A. -5
B. 0
C. 5
D. 4
E. -4
The mind is everything. What you think you become. -Lord Buddha



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by lolo » Sat Apr 04, 2009 5:19 am
E is correct.
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disagreeing with lolo :)

by kapsii » Sat Apr 04, 2009 7:29 am
Quadratic equations and geometry are not totally exclusive of each other, no elements of mathematics are totally independent which is why I find maths to beautiful & harmonious :) so, lets use Geometry to find the answer for this question...

Fact 1: Any quadratic equation corresponds to a parabola when plotted on a graph. (you can verify it by plotting values for the equation y = x ^ 2 + 6 x + 4)
So, to find the least value of the equation, is essentially asking us the lowest point on the Y axis for the parabola, looking at the quadratic equation, we know it is a U shaped parabola.

Fact 2: At the lowest/highest point of a parabola, the slope will be 0, which for a quadratic equation of the form Ax^2 + Bx + C will be -B/2A.

Thus, the minimum value of the quadratic equation can be found out by substituting the value of X as -B/2A (or x = -6/2 = -3) in the original equation,
hence the answer will be -5.

Just for those curious, one can come to the same conclusion by using calculus (hint: Find the derivative of the quadratic function f(x) = <quadratic equation>)
Cheers,
Dubes
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Re: disagreeing with lolo :)

by sanju09 » Mon Apr 06, 2009 1:29 am
kapsii wrote:Quadratic equations and geometry are not totally exclusive of each other, no elements of mathematics are totally independent which is why I find maths to beautiful & harmonious :) so, lets use Geometry to find the answer for this question...

Fact 1: Any quadratic equation corresponds to a parabola when plotted on a graph. (you can verify it by plotting values for the equation y = x ^ 2 + 6 x + 4)
So, to find the least value of the equation, is essentially asking us the lowest point on the Y axis for the parabola, looking at the quadratic equation, we know it is a U shaped parabola.

Fact 2: At the lowest/highest point of a parabola, the slope will be 0, which for a quadratic equation of the form Ax^2 + Bx + C will be -B/2A.

Thus, the minimum value of the quadratic equation can be found out by substituting the value of X as -B/2A (or x = -6/2 = -3) in the original equation,
hence the answer will be -5.

Just for those curious, one can come to the same conclusion by using calculus (hint: Find the derivative of the quadratic function f(x) = <quadratic equation>)
or should we rewrite it as y = (x + 3) ^ 2 - 5, which will be minimum at x = -3 and hence y (min) = -5. :)
The mind is everything. What you think you become. -Lord Buddha



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by hmboy17 » Fri Apr 10, 2009 11:42 am
I suppose the question is not asking you to find the value of X. In fact is is asking the min. value of whole expression so if we say Y= x ^ 2 + 6 x + 4 then it's asking for which value Y would be minimum. To find the answer just put the answer choises in the eq. E (-4 would be answer).

You can solve this question through Maxima and minima also but loooking at the GMAT structure I am not using it.
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by aj5105 » Thu May 21, 2009 1:03 am
f(x) = x^2 + 6x + 4

Differentiating, we get

f'(x) = 2x + 6

This function will be minimum when x = -5

(A)
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by sanju09 » Thu May 21, 2009 1:45 am
aj5105 wrote:f(x) = x^2 + 6x + 4

Differentiating, we get

f'(x) = 2x + 6

This function will be minimum when x = -5

(A)
If we follow your way, which is not an ideal way on GMAT, then it will go like this rather:

f (x) = x^2 + 6 x + 4

Differentiating, we get

f' (x) = 2 x + 6

and for maxima or minima, f' (x) = 0

or 2 x + 6 = 0, or x = -3; and at x = -3

f" (x) = 2 > 0, so a minima exists at x = -3

hence the least value of f (x) = (-3)^2 + 6 (-3) + 4 = -5 or [spoiler]A[/spoiler].
The mind is everything. What you think you become. -Lord Buddha



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