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What are the zero(s) of the function f(x) = the quantity of 5 x squared minus 25 x, all over x?
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@mathstudent55
@xapproachesinfinity
\[\frac{5x^2 - 25x}{x}\] what common factor do you see..?
5x
ok... so I'll do that \[\frac{5x(x - 5)}{x}\] what common factor do you see in the numerator and denominator..?
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x
ok so that leaves 5(x -5) = 0 so what value of x makes the equation true..?
5, 0
only 5 and when x = 0 in the original equation you have a zero denominator, which is undefined so at x = 0 you have a point of discontinuity... so the solution is x = 5 which comes from the simplified version of the original equation. hope it makes sense.
It does! Thank-you:)
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