Algebra parabola help!
I need help with plotting parabolas Y=x^2-5x+4
Y=x^2-4x
Y=-x^2+6x-8
Please provide the steps please.. i am very desperate I need a solid foundation of this :O thanks again!
November 4, 2010
Algebra parabola help!
I need help with plotting parabolas Y=x^2-5x+4
Y=x^2-4x
Y=-x^2+6x-8
Please provide the steps please.. i am very desperate I need a solid foundation of this :O thanks again!
7 Comments • Newest first
Ok, I got this guys thanks a bunch!
Y=x^2-5x+4
Firstly, factor into term a(x-p)^2+q
p is your horizontal translation. -2 is translation right by 2.
q is your vertical translation. For this one can't remember if + is up or down.
a is your stretch factor of 1/a.
To plot, use the original parabola then apply transformations/ translations. Original parabola start with vertex at 0,0.
If this is a pre-calculus course, find the P value and the vertex through: (X sub1 - X sub2)^2= 4P(Y sub1 - Ysub2)
Both the sub2s are the coordinates of the vertex.
Or just substitute in numbers.. ;o
-b/(2a) to find the x coordinate of the vertex. Plug that value into the equation of the parabola to find the y value. Plot that.
Next you should find the location 2 or 3 numbers above and below the x value of the vertex. So let's say the vertex is 3. Plug in 1, 2, 4, and 5 into the equation to find their y coordinates. Plot each point. Then connect them.
I wish people could more helpful
I need help as well. After i find the vertex I'm so lost!
[quote=xxsasuke20xx]Put that into your graphing calculator?[/quote]
No graphing calculators allowed.. or else I would basically not ask you guys