## neverforgetvivistee 4 years ago can someone please help me in understanding graphing quadratic inequalities? look at this (answers are below) http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Graphing%20Quadratic%20Inequalities.pdf what does the solid/shaded line mean? Why is it shaded in and some are shaded out? I know how to get the coordinates by plugging in random x values, but I don't know about anything else.

1. asnaseer

ok, in general, lets say you have some function in the form:\[y=f(x)\]and you have some inequality like:\[y>=4\]

2. asnaseer

now, on the line \(y=f(x)\) you know the value of 'y' will EQUAL the value of 'f(x)', so we draw the curve for y=f(x) as a solid line to indicate that we need to include this region.

3. asnaseer

so now we need to consider the line y=4. this will be a horizontal line which passes through y=4.

4. asnaseer

the horizontal line could intersect the curve at some points. e.g.: |dw:1322436287068:dw|

5. asnaseer

now since our inequality is y>=4, we need to draw a solid line at the places where y=4.

6. asnaseer

|dw:1322436426127:dw|

7. neverforgetvivistee

8. asnaseer

sorry - my explanation went a bit hay wire above!

9. asnaseer

what I should have said is lets say we have an inequality of the form y>=f(x)

10. neverforgetvivistee

lol it's okay take your time :)

11. asnaseer

now, on the curve y=f(x), we know it satisfies the inequality - so we make the line solid

12. asnaseer

below the curve y=f(x) does NOT satisfy the inequality as there we have y<f(x)

13. asnaseer

above the curve y=f(x) DOES satisfy the inequality, so we include that region by shading it in

14. neverforgetvivistee

for shading it in, how do you know it satisfies the inequality?

15. asnaseer

|dw:1322436711384:dw|

16. asnaseer

does the diagram make sense?

17. neverforgetvivistee

not really :(

18. asnaseer

I have taken some point x=x1 (which is represented by the vertical line) where this vertical line crosses the curve y=f(x), we know y=f(x1) above that point of intersection, we know y>f(x1) below that point of intersection, we know y<f(x1)

19. neverforgetvivistee

thanks :)

20. asnaseer

np - sorry for the confusion at the beginning :-)