## aussy123 2 years ago How do you graph this x2- 3x + 2<= 0

1. aussy123

$x^2+3x+2\le0$

2. dopeboz

wow!!!! i wouldnt know sorry

3. aussy123

its ok

4. dopeboz

k. but anyways i think your cute tho.

5. helder_edwin

first solve it as if it were an equation $\large x^2-3x+2=0$

6. myko

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|

7. aussy123

oh so I dont have to solve for anything

8. myko

|dw:1351782141285:dw|

9. helder_edwin

no it is not. @myko !!!!!

10. helder_edwin

the inequality has only ONE variable. its graph CANNOT bidimensional.

11. helder_edwin

anyways. myko just gave the answer $\large 0=x^2-3x+2=(x-1)(x-2)$ so x=1 or x=2

12. myko

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

13. aussy123

ok im a bit confused

14. myko

i gave the cross points to be able to grapfh it. The answer is the part under x.-axis

15. myko

@aussy123 don't be confused by @helder_edwin . The graph you looking for is: |dw:1351782419649:dw|

16. aussy123

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

17. myko

yw

18. helder_edwin

again. the problem says: graph $$x^2-3x+2\leq0$$ it does NOR say graph $$x^2-3x+2\leq y$$.

19. helder_edwin

sorry *NOT

20. myko

problem says: $y \leq 0$

21. myko

you wrong pal

22. myko

How do you GRAPH this x2- 3x + 2<= 0

23. helder_edwin

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.

24. myko

x2- 3x + 2 represents values of y, dude.

25. myko

it is one dimentional, :)

26. myko

y is DEPENDENT

27. myko

@helder_edwin

28. helder_edwin

but it was never there. u r not graphing a function "dude".