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| EAfreak |
 Posted: Tue Apr 25, 2006 5:35 pm Post subject: Algebra 2 (Help me for the love of God!) |
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Forum Freshman
Joined: 24 Apr 2006
Posts: 6
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I need some help on about five questions... The first two I need to find a quadratic equation that has the give roots:
The first question: 3/4, 9/2
The second question: -0.25, 0.25
Then next three I need to write each equation in the form f(x) = a(x-h)^2+k (unless it's not already in that form). I need to name the vertex, axis of symmetry, and direction of opening for the graph of eahc quadratic function. Then I need to graph it.
The third question is: f(x)=a(x-5)^2-7
The fourth question is: f(x)= -9^2+54x-8
The fifth (and final) question is: f(x)=0.25x^2-6x-16
Please Help!!  |
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| mitchellmckain |
| Posted: Wed Apr 26, 2006 7:04 am Post subject: Re: Algebra 2 (Help me for the love of God!) |
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Forum Cosmic Wizard
Joined: 06 Oct 2005
Posts: 2078
Location: Salt Lake City, UTAH, USA
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| EAfreak wrote: |
I need some help on about five questions... The first two I need to find a quadratic equation that has the give roots:
The first question: 3/4, 9/2
The second question: -0.25, 0.25
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Simplicity itself: let me give you an example 4, -1.45
I simply write (x-4)(x-(-1.45)) = 0
now if x is 4 or -1.45 I get 0
I can multiply this out using (a+b)(c+d)= ac+ad+bc+bd
x^2 - 2.55 x - 5.8 = 0
| EAfreak wrote: |
Then next three I need to write each equation in the form f(x) = a(x-h)^2+k (unless it's not already in that form). I need to name the vertex, axis of symmetry, and direction of opening for the graph of eahc quadratic function. Then I need to graph it.
The third question is: f(x)=a(x-5)^2-7
The fourth question is: f(x)= -9^2+54x-8
The fifth (and final) question is: f(x)=0.25x^2-6x-16
Please Help!!
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Again an example: f(x)=x^2 - 2.55 x - 5.8
The process is known as completing the square
and it uses (a-b)^2 = a^2 -2ab + b^2
where in our f(x), a^2 = x^2 and 2ab = 2.55 x
This means that a=x and b = 2.55/2 = 1.275
so (x-1.275)^2 = x^2 -2.55 x + 1.625625
So we take or f(x) and add (x-1.275)^2 and subtract x^2 -2.55 x + 1.625625 to get (x-1.275)^2 -5.8 -1.625625
Since what we added and subtracted are equal it is still f(x)
So f(x) = (x-1.275)^2 - 7.425625
SO the vertex of the parabola is at (1.275, -7.425625)
The axis of symmetry is x = 1.275
and the parabola opens upward because comparing
f(x) = (x-1.275)^2 - 7.425625 to f(x) = a(x-h)^2+k
we see that a = +1 which is positive
_________________
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| orca |
| Posted: Wed Jul 02, 2008 1:16 am Post subject: |
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Forum Freshman
Joined: 02 Jul 2008
Posts: 1
Location: new zealand
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For a quadratic equation of the form x^2 + bx + c = 0
The sum of its two roots is -b
The product of its two roots is c.
the first equation is:
x^2 -(3/4 + 9/2) + (3/4)*(9/2) = 0
x^2 - (21/4)x + 27/8 = 0
8x^2 - 42x + 27 = 0
The second equation is
x^2 +(0.25 - 0.25)x +0.25*(-0.25) = 0
x^2 - 1/16= 0
16x^2 - 1 = 0
***************************************************
Write equations in the form f(x) = a(x-h)^2+k.
The fourth equation:
-9x^2+54x-8
= -9(x^2 - 6x + 8/9)
= -9[(x^2 - 6x + 9) - 9 + 8/9]
= -9[(x - 3)^2 - 9 + 8/9]
= -9(x - 3)^2 + 81 - 8
= -9(x - 3)^2 + 73
The fifth equation
0.25x^2-6x-16
= 0.25[x^2 - 24x - 64]
= 0.25[(x^2 - 24x + 144) - 144 -64]
= 0.25)[(x -12)^2 - 208]
= 0.25(x - 12)^2 - 52
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| serpicojr |
| Posted: Wed Jul 02, 2008 8:00 am Post subject: |
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Forum Professor
Joined: 17 Jul 2007
Posts: 1128
Location: JRZ
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| orca: On this forum, we prefer not to answer homework questions, but rather we attempt to help people by providing examples (as mitchellmckain did) or by asking leading questions. Of course, we're excited to have you on the math forum--just keep this in mind for the future! |
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