Homework Help: Algebra II

Posted by Emily on Wednesday, December 17, 2008 at 8:49pm.

I honestly forgot how to do these type of problems. Can someone help me! Thanks!!

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24. ( 9x^2 - 4 )/( 3x^2 - 10x - 8 )

31. 2x^2 + 5 = 12x

39. Find the maximum value of g(x) = -4x^2 + 16x - 15

42. Graph y-9 = -2(x+3)^2

Algebra II - Damon, Wednesday, December 17, 2008 at 9:02pm
( 9x^2 - 4 )/( 3x^2 - 10x - 8 )
well the numerator is difference of two squares so factor
(3x-2)(3x+2)
now I bet one of those is a factor of the denominator
(3x - 2)(x+4) no
(3x+2)(x-4) yes
so we really have
(3x+2)(3x-2) / (3x+2)(x-4)
which is
(3x-2) / (x-4)
Algebra II - Damon, Wednesday, December 17, 2008 at 9:07pm
2 x^2 - 12 x + 5 = 0
I can not factor that, sure it is not 11 x?
If not a typo, use quadratic equation
x = [ -b +/- sqrt(b^2-4ac)] / 2a
Algebra II - Damon, Wednesday, December 17, 2008 at 9:17pm
g(x) = -4x^2 + 16x - 15
parabola
g is - for bi + or - x so upside down (sheds water)
Find the vertex.
4 x^2 - 16 x + 15 = -g
x^2 - 4 x = (-g-15)/4
x^2 - 4 x + 4 = -g/4 - 15/4 + 16/4
(x-2)^2 = (1/4) (-g +1)
AT VERTEX x = 2, g = +1
so max at (2,1)
Algebra II - Damon, Wednesday, December 17, 2008 at 9:22pm
y-9 = -2(x+3)^2
pretty much the same as the last one
same both sides of x = -3 so that is the vertical axis of symmetry
vertex at (-3,9)
when x gets big, y gets small, so sheds water
when y = 0
9/2 = (x+3)^2
x + 3 = +/- 3 /sqrt 2
x = -3 +/- 3/sqrt 2 is where he zeros are
Algebra II - Damon, Wednesday, December 17, 2008 at 9:23pm
Just brute force think your way through these problems !!!
Algebra II - Emily, Thursday, December 18, 2008 at 12:30am
OMG you are FREAKIN AWESOME for helping me out with these problems!!!!!! With your thorough explanations it all came back to me I thank you SOOOOOO MUCH!!!!!!
Algebra II - Mike, Friday, March 30, 2012 at 1:54am
6x^4-15x^3+10x^2-10x+4 divided by 3x^2+2
Please help!!!

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