2nd to last question
A vertical asymptote is where the denominator is equal to zero.
@Xeph @linh412986 @ehuman
So, set the denominator equal to zero and solve for the x-values that make vertical asymptotes.
@JoannaBlackwelder is right. Look at the denominator and set it to zero, and solve for x
Nothing more to say, Joanna is right. Try to think a little bit. @hannahmichelleee
You will need to factor the quadratic into binomials.
when I set it to 0 and solve I get -4,2 on my graphing calc.
Me too. :)
Good, -4 and 2 are both correct answers
so thats it!? :D I'm done with that equation?
you shouldn't need a calculator for this problem. the denominator is \[x^2+2x-8\]set it to zero,\[x^2+2x-8=0\]then we factor. numbers that multiply to -8 and add to 2? factors of 8 are 1,2,4,8 so we are looking at\[\pm 2,\pm 4\]but they add to +2, so we need +4, -2 so we have \[(x+4)(x-2)=0\]solving for x is trivial then, you have 2 solutions
thank you all!!!
Yup, that 's it!
Join our real-time social learning platform and learn together with your friends!