the minimum value

E is correct.

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>)

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.

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.

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

Differentiating, we get

f'(x) = 2x + 6

This function will be minimum when x = -5

(A)

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].