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Mathematics 15 Online
OpenStudy (anonymous):

(x-4)/(x^2-5x+4) Find the values of x at which there are discontinuities. Simplify the fraction. Based on the simplified fraction, where is there a vertical asymptote, and where is there a removable discontinuity?

OpenStudy (zzr0ck3r):

you need to find the zeros of x^2-5x+4 x^2-x-4x+4=0 x(x-1)-4(x-1)=0 (x-4)(x-1)=0 so x=4 and x=1 are the points of discontinuity

OpenStudy (anonymous):

thank you

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