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Mathematics 13 Online
OpenStudy (anonymous):

Is 1 the x-intercept for: f(x)=x^3-1/x^2-x

jimthompson5910 (jim_thompson5910):

Hint: x^3-1 = (x-1)(x^2 + x + 1)

OpenStudy (anonymous):

so there is no x-intercept

jimthompson5910 (jim_thompson5910):

what makes you say that

OpenStudy (anonymous):

x can never = 0

jimthompson5910 (jim_thompson5910):

true, but that doesn't mean we won't have x-intercepts

OpenStudy (anonymous):

=/

jimthompson5910 (jim_thompson5910):

can you simplify (x^3-1)/(x^2-x)

OpenStudy (anonymous):

would x^2 cancel out and be left with x-1/-x?

jimthompson5910 (jim_thompson5910):

no

jimthompson5910 (jim_thompson5910):

x^3-1 = (x-1)(x^2 + x + 1) x^2-x = x(x-1)

jimthompson5910 (jim_thompson5910):

what cancels?

OpenStudy (anonymous):

x+1 and x-1

jimthompson5910 (jim_thompson5910):

you mean the two x-1 terms right?

OpenStudy (anonymous):

correct

jimthompson5910 (jim_thompson5910):

so you're left with x^2 + x + 1 ----------- x

jimthompson5910 (jim_thompson5910):

set that whole thing equal to 0 and solve for x

OpenStudy (anonymous):

i see...wouldn't this also mean there would be a Rational discontinuity?

jimthompson5910 (jim_thompson5910):

there is a discontinuity and it's at x = 0 since x can't be 0

OpenStudy (anonymous):

would i have to use the quadratic formula on x^2+x+1 in order to find an x-intercept?

jimthompson5910 (jim_thompson5910):

yes, that's the most direct way to do so (since factoring involves guessing)

OpenStudy (anonymous):

this would give me an imaginary number though correct?

jimthompson5910 (jim_thompson5910):

that is correct

jimthompson5910 (jim_thompson5910):

so there are no real x intercepts

OpenStudy (anonymous):

awesome! thanks for the help!

jimthompson5910 (jim_thompson5910):

np

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