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I have a rational function: $y = \frac{x^2 - 3x}{x^2 + 2x - 48}$

The explanation I was given to find the x-intercepts was: "let y = 0, and solve for x. Basically you just set the numerator of the fraction equal to 0 and factor."

What does "basically" mean? Is this always the case? Do you simply set the numerator expression equal to zero, and always ignore the contents of the denominator? If so, why do you ignore the contents of the denominator?

saulspatz
  • 53,131
  • If you start with $0 = \frac{a}{b}$, you multiply both sides by $b$, you get $0 = a$. This is just because anything times zero is zero. – Nick May 12 '21 at 18:46
  • For this rational function, $$ f(x) \ = \ \frac{x^2 - 3x}{x^2 + 2x - 48} \ = \ \frac{x·(x-3)}{(x+8)·(x-6)} \ \ . $$ Since there are no common factors between the numerator and the denominator, the values of $ \ x \ $ which make the numerator equal to zero will not also make the denominator zero, so the instruction you were given will get you the $ \ x-$ intercepts. If there were such a common factor, the function has a "hole" in its domain at that value of $ \ x \ . $ If the denominator cannot be factored using real numbers, the function is always defined and the simple instruction applies. –  Aug 02 '21 at 07:42

1 Answers1

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You can't just ignore the denominator. After following the suggested procedure, you have to check whether the point you found is also a zero of the denominator. If so, $y$ is not defined at the point in question.

$x$ has no chance of being a zero unless it's a zero of the numerator, but that's not the whole story.

saulspatz
  • 53,131
  • So in other words I factor the expression in the numerator to determine the possible x-intercepts. Then plug in each possible x-intercept into the denominator to ensure that it does not equal 0 in the denominator, to omit any possible x-intercepts that would make the denominator 0? – VerySeriousSoftwareEndeavours May 12 '21 at 19:22
  • Yes, that's correct. – saulspatz May 14 '21 at 23:04