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wow!!!! i wouldnt know sorry
k. but anyways i think your cute tho.
first solve it as if it were an equation \[ \large x^2-3x+2=0 \]
it's a parabola. To find where it crosses x-axis, write it like this: x^2-3x+2=(x-1)(x-2)=y so it will cross x-axis at x=1 and x=2 since you lookiing for points that are less than 0 it will be the part that is between this two points: |dw:1351781925843:dw|
oh so I dont have to solve for anything
no it is not. @myko !!!!!
the inequality has only ONE variable. its graph CANNOT bidimensional.
anyways. myko just gave the answer \[ \large 0=x^2-3x+2=(x-1)(x-2) \] so x=1 or x=2
look my last drawing x^2-3x+2=y is a function with range (vertex, infinity) it is less or equal 0 in the part that is under x-axis. That's what i drowed
ok im a bit confused
i gave the cross points to be able to grapfh it. The answer is the part under x.-axis
@aussy123 don't be confused by @helder_edwin . The graph you looking for is: |dw:1351782419649:dw|
Oh I see , @myko , I had to reread your first comment to get it, thanks because my book is giving me alot of confusing stuff
again. the problem says: graph \(x^2-3x+2\leq0\) it does NOR say graph \(x^2-3x+2\leq y\).
problem says: \[y \leq 0\]
you wrong pal
How do you GRAPH this x2- 3x + 2<= 0
it is a one-dimensional problem (one variable) but u turned it into a two-dimensional problem by adding a "y" that was never there.
x2- 3x + 2 represents values of y, dude.
it is one dimentional, :)
y is DEPENDENT
but it was never there. u r not graphing a function "dude".