calculus
posted by Unknown .
The function f is given by the formula f(x)=6x^3+17x^2+4x+21/x+3
when x<–3 and by the formula
f(x)=2x^2–4x+a
What value must be chosen for a in order to make this function continuous at 3?

mmmh, took me a while to understand what the question is asking.
First of all you will need brackets:
f(x) = (6x^3+17x^2+4x+21)/(x+3)
= (6x^2  x + 7) , where x≠ 3
= (x+1)(6x7)
we want this to be the same as
2x^2  4x + a
If 6x^2  x + 7 = 2x^2  4x + a
4x^2 + 3x + 7 = a
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The function f is given by the formula f(x)=6x^3+17x^2+4x+21/x+3 when x<–3 and by the formula f(x)=2x^2–4x+a What value must be chosen for a in order to make this function continuous at 3?