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What is the minimum value of the function y = x^2 - 4x - 5?

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What is the minimum value of the function y = x^2 - 4x - 5? [#permalink]

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New post 19 Jan 2012, 17:08
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What is the minimum value of the function y = x^2 - 4x - 5?

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Re: What is the minimum value of the function y = x^2 - 4x - 5? [#permalink]

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Splendidgirl666 wrote:
What is the minimum value of the function y=x^2-4x-5?
Can someone explain how to approach this one?


We can solve this one with parabola function approach, though I think the following will be easier:

\(y=x^2-4x-5=(x^2-4x+4)-9=(x-2)^2-9\), now as \((x-2)^2\geq{0}\) (so min value is 0) then the minimum value of the function is 0-9=-9.

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Re: What is the minimum value of the function y = x^2 - 4x - 5? [#permalink]

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New post 20 Jan 2012, 23:11
If you are at all familiar with calculus taking the derivative setting it = 0 and plugging in values may be easier here. Just remember that if you only get one value you need a second derivative test to check whether it is a min or a max.

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Re: What is the minimum value of the function y = x^2 - 4x - 5? [#permalink]

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New post 21 Jan 2012, 02:32
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joeshmo wrote:
If you are at all familiar with calculus taking the derivative setting it = 0 and plugging in values may be easier here. Just remember that if you only get one value you need a second derivative test to check whether it is a min or a max.


Yes, you can use derivative for this: \(y'=2x-4\) --> \(2x-4=0\) --> \(x=2\) --> substituting this value in the original expression we'll get: \(x^2-4x-5=4-8-5=-9\).

As for the second part of your post: as the coefficient of x^2 in x^2-4x-5 is positive then this function will have ONLY minimum value and not maximum, so no need for further checking (y=x^2-4x-5 is an upward parabola and the minimum of this function is y-coordinate of its vertex). Check for more on parabolas in Coordinate Geometry chapter of Math Book: math-coordinate-geometry-87652.html

Also note that derivative of a function is not tested on the GMAT.
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Re: What is the minimum value of the function y = x^2 - 4x - 5? [#permalink]

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New post 16 Jan 2015, 08:19
y=x^2-4x-5
1) -> (x-5)*(x+1)
so y=0 for x=5 and x=-1
the function states that the lowest point got to be between these 2 points (5/-1). it should be the middle on the line,
so in -1 0 1 2 3 4 5

is this a valid method Bunuel
thanks in advance

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Re: What is the minimum value of the function y = x^2 - 4x - 5? [#permalink]

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New post 16 Jan 2015, 08:36
Use the derivative test

Lets take an example. y=x^2-4x-5

dy/dx = 2x - 4....First Derivative

Equate it to 0 to get x value.....2x - 4 = 0 , x = 2.

Your question ends here.

But examiner can ask the minimma or maximuma for quadratic
How to find?

Double Derivate it

d2y/dx2 = 2............Second Derivative is +ve so maxima.

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Re: What is the minimum value of the function y = x^2 - 4x - 5? [#permalink]

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New post 16 Jan 2015, 14:16
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Hi All,

The GMAT will NEVER require that you know Calculus/Derivatives to get to the correct answer to a Quant question.

IF the original poster had included the 5 answer choices, then we'd have those to use for reference (and we'd likely need to do LESS work to get to the solution). As it is, we can find the correct answer with a little bit of "brute force" and some Number Properties.

We're given X^2 - 4X - 5 and asked for the MINIMUM value that can be generated from this equation.

Here are the Number Properties worth noting:
1) X^2 will either be positive or zero.
2) If X is NEGATIVE, then -4X becomes positive (which INCREASES the value of the calculation and that is NOT what we want)
3) Since we're dealing with X^2, the graph of this equation will "curve" somehow (it won't be a straight line)

To minimize the outcome, X WILL NOT be negative....so let's look for a pattern in 0 and the positive possibilities:

If X=0, the result is 0 - 0 - 5 = -5
If X=1, the result is 1 - 4 - 5 = -8
If X=2 the result is 4 - 8 - 5 = -9
If X=3 the result is 9 - 12 - 5 = -8
If X=4 the result is 16 - 16 - 5 = -5

From this, the answer appears to be -9. The answer choices, if we had them, would help to confirm this (as we could eliminate answers that were "too big" or not possible).

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Re: What is the minimum value of the function y = x^2 - 4x - 5? [#permalink]

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New post 20 Nov 2015, 06:55
Hi,
I have another approach.
Though Derivative is not required for GMAT, I have seen some simple problems that can be solved by the derivative of a function.
Our function here is y=x^2-4x-5
Derivative of this function will be 2x-4 which is equal to 0
2x=4
x=2
let us plug x=2 in y.
we will have y=2^2-4*2-5=4-8-5=-9

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What is the minimum value of the function y = x^2 - 4x - 5? [#permalink]

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Splendidgirl666 wrote:
What is the minimum value of the function y = x^2 - 4x - 5?

I would just find the vertex of the parabola that corresponds with the graph of the quadratic function.

The y-coordinate of the vertex is the minimum value of a parabola that, as here, opens upward (positive coefficient for \(a\), squared term is x).

I do not like completing squares in order to put the quadratic function in vertex form. Technically, and especially when the function is in vertex form, the vertex is (h,k). I'll use x and y here.

Just two steps: 1) find the axis of symmetry and x-coordinate of vertex, then 2) plug that value into the function to find y-coordinate, which is this function's minimum value.

Use \(\frac{-b}{2a}\) to find the axis of symmetry and x-coordinate of the vertex.

\(x =\frac{-b}{2a} =\frac{-(-4)}{2} = 2\)

\(x\) = 2

Evaluate y at x (plug x = 2 into the function) to find y-coordinate:

\(y = x^2 - 4x - 5\)
\(y = 2^2 - 4(2) - 5\)
\(y = -9\)

The minimum value of an upward curving parabola is the y-coordinate of the vertex.

Hence the minimum value of the function \(y = x^2 - 4x - 5\) is -9. y will never be less than -9.

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Re: What is the minimum value of the function y = x^2 - 4x - 5? [#permalink]

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New post 24 Oct 2017, 21:43
We could find the Y coordinate directly
axˆ2+bx+c
Y=c-(b^2/4a)
y=xˆ2−4x−5
a=1. b=-4 & c=-5
=-5-((-4)ˆ2/(4*1))
=-5-(16/4)
=-5-4=-9

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Re: What is the minimum value of the function y = x^2 - 4x - 5? [#permalink]

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New post 24 Oct 2017, 21:45
We could find the Y coordinate directly
Y=\(ax^2\)+bx+c
Y=c-(\(b^2\)/4a)
y=\(x^2\)−4x−5
a=1. b=-4 & c=-5
=-5-(\((-4)^2\)/(4*1))
=-5-(16/4)
=-5-4=-9

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Re: What is the minimum value of the function y = x^2 - 4x - 5?   [#permalink] 24 Oct 2017, 21:45
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