Standard Form Question

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blitzen

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Jan 7, 2011
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How would I go about putting f(x) = 2x^2 - 4x + 7 into standard form?

I thought I knew how to do it, but I ended up trying to factor it. I feel like a simpleton for having to ask this because I used to know how.
 
Factor? Did you try "Completing the Square"? Of course, it depends on exactly what "Standard Form" means.
 
This one is weird to factor. Normally I would say that I need two numbers adding to -4 and multiplying to 14, right? I cannot come up with said numbers.
 
blitzen said:
This one is weird to factor. Normally I would say that I need two numbers adding to -4 and multiplying to 14, right? I cannot come up with said numbers.
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I think we still need to know how "standard form" has been defined for you. Then we can proceed.
 
My professors question simply says to put it in standard form, with not definition given. I would imagine Ax + By = C
 
blitzen said:
I would imagine Ax + By = C

Whoa! You're heading way off-course, here. 8-)
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Do you know the difference between a linear function (i.e., first-degree polynomial function) and a quadratic function (i.e., second-degree polynomial function)?

I learned that Standard Form for a quadratic function is:

f(x) = A * (x - h)^2 + k

where A is the leading coefficient, and the ordered pair (h, k) represents the coordinates of the vertex point.

tkhunny suggested completing the square. That's how I would get the form above, too.

tkhunny asked you about completing the square; you have yet to answer that question.

 
Oh wow. I am going the wrong way, huh! That does ring a bell actually because my teacher had a saying for "h comma k." In addition, I need to find a vertex so that is definitely the route I need to go. As far as completing the square:

I have 2x^2 - 4x + 7.
2(x^2 - 2x + 7/2)
2[(x-1)^2 + 5/2]
2(x-1)^2 + 5 ??

Is this correct?
 
Re:

mmm4444bot said:
blitzen said:
2(x - 1)^2 + 5

Is this correct?

Yes, but report it using function notation.

f(x) = 2(x - 1)^2 + 5
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Click to expand...
So that makes makes my vertex (1, 5) right?
Also, I have to find the maximum and minimum. I do not have a max, do I? And my minimum would be (1,5)?
 
blitzen said:
So that makes makes [the] vertex (1, 5) right?

Correct.


[The function has no] max, [does it]?

There is no function maximum, unless they've specifically constrained the domain to something other than all Real numbers.


[The] minimum would be (1,5)?

No.

(1, 5) is an ordered pair of numbers. It represents the coordinates of the vertex point.

The function minimum is the smallest value of f(x) taken on over the entire domain.
Click to expand...
 
blitzen said:
I have 2x^2 - 4x + 7.
2(x^2 - 2x + 7/2)
2[(x-1)^2 + 5/2]
2(x-1)^2 + 5 ??

Is this correct?
Click to expand...
Actually, no. You have rather deliberately tried to confuse yourself. Leave out the constant at first.

2x^2 - 4x + 7

2*(x^2 - 2x) + 7

2*(x^2 - 2x + ___) + 7 - 2*(___)

(2/2)^2 = 1^2 = 1

2*(x^2 - 2x + 1) + 7 - 2(1)

2(x - 1)^2 + 5

In this case (it will not always be so), we also managed to avoid fractions. Fractions are not evil, but life is a little easier with integers.
 
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