## creighton 2 years ago Need help on a Algebra 2 Honors Question!

1. creighton

4.f(x)= -1 if x ≤ -6 5 if -6 < x ≤ -2 2x + 9 if x > -2 (1 Point)

2. creighton

I also have part of a table filled out

3. creighton

4. creighton

I just don't understand what I am supposed to do for the first 4 columns.

5. latremese40

if i was good at this ill help but me im not that good

6. creighton

I just know about the las column never did any like the first two. I could post other I have already done.

7. creighton

8. latremese40

like what r u suppose to do?

9. creighton

plug in numbers for coordinates and graph it.

10. creighton

Example fx=2x+9 if x > -2

11. creighton

Next you would chose a value for x that is greater than -2 which is 0

12. latremese40

oh i have no idea but i can ask my bro to help u

13. creighton

so now its 2(0)+9 which gives us 9 if we solve it and that is our y value

14. latremese40

ok...

15. creighton

giving us the coordinates for a point at (0,9) if we do a few more we get a graph for two of the columns

16. creighton

I just don't now what to do if there is no x value for the equation.

17. latremese40

yeahh im confused on the same thin ur talkin bout o dont get it either

18. Blacksteel

This function works just like any other in the other two sections. For -6 < x < -2, the function is just y = 5, so for any x value in that range you'll get 5 as your result. Similarly, for any x < -6, y = 1.

19. creighton

so any number would be -1 for the first two columns

20. Blacksteel

For the first two columns, you'll want to use different x values, but the y value will always be -1. So for example, the first two columns might look like: $\left[ -6 | -1 \right]$$\left[ -7 | -1 \right]$$\left[ -8 | -1 \right]$

21. creighton

and the graph would be a horizontal line on y=-1 right

22. Blacksteel

Yes.

23. creighton

thanks a lot I relly needed help understanding this and this is the only question that did those with nothing in the lesson that talked about it. Thanks again

24. Blacksteel

No problem, feel free to message me if you have any other questions.