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f(x) = 4x^2 + 28x + 49 for all x

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f(x) = 4x^2 + 28x + 49 for all x [#permalink] New post 07 Dec 2017, 03:11
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f(x) = 4x^2 + 28x + 49 for all x

Quantity A
Quantity B
The number b such that f(b) is the minimum value of f
-3


A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal
D. The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

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Re: f(x) = 4x^2 + 28x + 49 for all x [#permalink] New post 02 Feb 2018, 20:57
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This problem is juicy. Essentially we must find the minimum for the quadratic equation. What's the minimum? As you know, a quadratic will produce a line with a single curve. The minimum is just the lowest point on the curve. This also, btw, implies that the curve is opening upwards since a downwards opening curve would go down to infinity. There are a ton of ways to approach this problem. Let's talk about the worst ways first.

Firstly, you may be tempted to just plug in -3 into the equation and see what happens. Here's what'll happen: you'll waste a decent amount of time multiplying some sloppy numbers by -3 and then get another number (it's 1) and then what? What does that mean? Is that the minimum? The only way to know is to plug in other numbers and see whether they produce a smaller value or not. This could be pretty time-consuming.


You might also consider using the quadratic formula. Remember that one? It's simply (-b ± √(b^2-4ac))/2a. So simple! That's sarcasm. Listen: you are never required to use the quadratic formula on the GRE. You can use it if you'd like, but it's time-consuming and complicated, two things we're against.


However, if you do have the roots, then we can get the value of x that would get us the minimum. After all, quadratic graphs are always symmetrical, so the minimum or maximum will always fall exactly halfway between the two roots!

If you don't want to deal with the quadratic formula, you can factor out the quadratic. When there's a multiplier on the squared variable, this isn't that easy. However, what if we divided the whole thing by 4? Would that work? Basically, yes. The equation would be a different equation, but it would be the exact same shaped, just vertically scrunched by a factor of 4. The key point though is that it would have the same two roots. Once we have those, we can figure out what value of x is between them and we're home free.

So if we divide the whole thing by 4 let's write it this way: x^2 + (14/2)x + 49/4. Now we can factor as we usually do: what times what makes 49/4 but added makes 28/4? Seems tough but there's a clue in the fact that 49/4 is a perfect square. So let's try its square root, which is 7/2. So 7/2 + 7/2 does indeed make 14/2, and 7/2 squared makes 49/4. That means that the scrunched version of the function, factored, is (x + 7/2)(x + 7/2). So it's only got one root, which is -7/2, or -3.5. When a quadratic only has one root, that's because it's only touching the x-axis once, which means that point is it's minimum or maximum. Thus, the minimum of this equation must occur when x = -3.5, so the answer is B. Whew!

EPILOGUE:
Honestly, the easiest way to solve this is to just take the derivative of the equation and then set it equal to zero. This would give you 8x + 28 = 0 so x = -3.5. But of course calculus is never required on the GRE. If you already know it, great. But you certainly don't need to learn calculus just for the GRE. Not an efficient use of your time.
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Re: f(x) = 4x^2 + 28x + 49 for all x [#permalink] New post 03 Feb 2018, 03:46
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Also another method (if you already know calculus) is:
f'(x)= 8x+28
equate it to zero:
8x+28=0 --> x=-7/2;
f'(x)=8; @ x=-7/2; f"(x)= 8 > 0 => f(x) is minimum @ x=-7/2.
So b is -7/2 => -3.5<-3

Answer is B.
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Re: f(x) = 4x^2 + 28x + 49 for all x [#permalink] New post 27 Apr 2018, 03:57
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There is a equation to find the vertex of parabola (x,y). A parabola is a u-shaped curve where the vertex is either the minimum point or the maximum point of the parabola, in this instance the vertex is the minimum point.
The method to find the value of y is a bit complicated however finding x is comparatively easier.
Here we just need to find the x value because at this point the value of y is also minimum.

the x point of the vertex of the parabola is \frac{-b}{2a} in the quadratic eqn of the form ax^2 + bx + c if a is +ve vertex is the minimum point else if a is -ve vertex is the maximum point
in the given problem 4x^2 + 28x + 49
b = 28 hence -b = -28
2a = 2*4 = 8
therefore x = -28/8 = -3.5
since -3.5<-3
option B
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Re: f(x) = 4x^2 + 28x + 49 for all x   [#permalink] 27 Apr 2018, 03:57
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