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DO you have any idea
Not not really. I stink at this stuff. Sorry :(
Do you know?
yep hold on im working the problem out :)
Factoring polynomials never goes away. You have to factor numerator and denominator. x^3+x^2-4x-4 can be factored "by grouping" as X^3+x^2-4x-4=(x^3-4x)+(x^2-4)+(x^2-4)=(x+1)*(x^2-4) It can be factored further because the difference of squares x^2-4=x^2-2^2 factors as (x-2)*(x+2) x^2+2x-3=(x+3)*(x-1)
oh and the function is not defined (it does not exist) for x=1 and for x=-3 because the denominator is zero for those values of x .
Hope this helps :) have a colorful day
Sorry my computer wasnt working
so would the answer be x=1 and x=-3
x=1 and x=-3 would result in a division by zero, and as you know that is a "no-no" and would result in a discontinuity.
Yes, that is what the problem is seeking.
so that is the answer
And what about this one Describe the vertical asymptote and hole for the graph of x^2+x-6/ x^2-9
a division by zero would occur a x=3
so is it x=3 and x=-3
I am not sure what occurs when x=-3 but x=3 and x=-3 is probably your best choice.
oh ok thanks... you know that question i sent you
and you asked if there was a slope
here is the graph
what is that...
did you get the graph
I believe the 3rd choice may be the best, still not sure about the
about the x = -3
oh ok... and what about the question with the graph
I also don't know the answer to the link graph. Not too much help today.
Regarding the graph, you do have a y intercept at x=-3. It is a function as a vertical line will not intercept the y values twice. It is a non-linear function.
It is a direct variation y=3x
it want me to write an equation
its asking for an equation
and I provided one for you. y=3x
I gotta run for now (lunch)
oh ok thanks!