For what value of 'k' is the function continuous at x=-3?
f(x)=(2x^2+5x−3)/(x^2−9) ,x≠−3
f(x)=k ,x=−3

Question

For what value of 'k' is the function continuous at x=-3?
f(x)=(2x^2+5x−3)/(x^2−9) ,x≠−3
f(x)=k ,x=−3

anonymous · Answer

allow me to help you with this problem

anonymous · Answer

Thanks, i don't know how to start it

anonymous · Answer

so first begin by pluging in -3

freckles · Answer

you need
$$f(-3)=\lim_{x \rightarrow -3}f(x)$$

anonymous · Answer

because that results in a 0 in the denominator, you are going to have to use another method. limits just as freckles has pointed out.

anonymous · Answer

in order to take the limit, lets factor both the top and bottom

anonymous · Answer

OK

anonymous · Answer

Here is the top
$$(2*x-1)(x+3)$$

and the bot would be
???

anonymous · Answer

(x+3)(x-3)?

anonymous · Answer

correct, so now what can you do with this fraction?

$$\frac{ (2*x-1)(x+3) }{ (x+3)(x-3) }$$

anonymous · Answer

cross off the (x+3) on the top and bottom and get $$\frac{ (2x-1) }{ (x-3) }$$

anonymous · Answer

would i then plug -3 in for x?

anonymous · Answer

correct!

anonymous · Answer

Ok thank so so much!! i got 7/6 for the answer

anonymous · Answer

that is what I have recieved. Glad i could be of help!