## anonymous 4 years ago i have 2 equations (4x3+24x2+7x+42)/(x+6)and(2x2+6x+a) the first formula has an x<-6 and the 2nd has x</=-6 and i have to find which value would make it continuous. i set them equal to each and plugged in -6 in for the x and solved for a getting -36 which is wrong

1. anonymous

factor and cancel to get $4x^2+7$

2. anonymous

replace x by 6 and get $4\times 6^2+7=189$

3. anonymous

so my algebra is wrong but my method is right?

4. anonymous

what you then have to do, now that you know the limit of the first expression is equal to 189, is replace x by 6 in the second expression, set it also equal to 189 and solve for "a"

5. anonymous

there is only 1 a though

6. anonymous

and that is in the 2nd equation

7. anonymous

right. the second expression is $2x^2+6x+a$ and you need to to match up to the first expression when x =6

8. anonymous

you know the limit of the first expression at 6 is 186 so write $2\times 6^2+6\times 6+a=189$ and solve for "a"

9. anonymous

sorry i meant "189"

10. anonymous

oh ok that makes sense

11. anonymous

i get $108+a=189$ and therefore $a=81$ but don't take my word for it

12. anonymous

it says 81 is in incorrect

13. anonymous

x had to equal -6 not 6

14. anonymous

x=-7, a=147 x=-8, a=183 and so on