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
factor and cancel to get \[4x^2+7\]
replace x by 6 and get \[4\times 6^2+7=189\]
so my algebra is wrong but my method is right?
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"
there is only 1 a though
and that is in the 2nd equation
right. the second expression is \[2x^2+6x+a\] and you need to to match up to the first expression when x =6
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"
sorry i meant "189"
oh ok that makes sense
i get \[108+a=189\] and therefore \[a=81\] but don't take my word for it
it says 81 is in incorrect
x had to equal -6 not 6
x=-7, a=147 x=-8, a=183 and so on
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