Let f ( x ) = x 2 + 1 x 2 − 9 So to find the y intercept I take f ( 0 ) correct? So when I substitute 0 for x I got − 1 9 so is the y intercept ( 0 , − 1 9 )Also to find the x intercept I set the numerator equal to 0. So then I got x 2 + 1 = 0 but wouldn't that make x 2 = − 1 which is imaginary?

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Let   f  (  x  )  =                              x          2                +        1                              x          2                −        9            So to find the y intercept I take   f  (  0  ) correct? So when I substitute   0 for   x I got   −      1    9   so is the   y intercept   (  0  ,  −      1    9    )Also to find the   x intercept I set the numerator equal to   0. So then I got       x    2    +  1  =  0 but wouldn&#039;t that make       x    2    =  −  1 which is imaginary?

Cristian Hamilton · Accepted Answer

Yes. This is correct. To find the y intercept of a function,   f  (  x  ), find   f  (  0  ).Your approach for the x intercept is also correct. Assuming that you are only trying to find intercepts in the real numbers, taking       x    2    +  1  =  0 means that there are no   x intercepts.You calculated your derivative incorrectly. The derivative of   f  (  x  ) is given by:            d      f              d      x        =            (              x        2            −      9      )      (      2      x      )      −      (              x        2            +      1      )      (      2      x      )              (              x        2            −      9              )        2              =            −      20      x              (              x        2            −      9              )        2            Setting             d      f              d      x        =  0 yields   0  =  −  20  x. However, you must also consider the points at which the derivative is undefined. To do this, solve   0  =      x    2    −  9. Finally, consider where             d      f              d      x       is positive and negative using various values.

Lydia Carey · Accepted Answer

You are correct about the y-intercept, though the terminology is a little ambiguous. It can mean the intersection point on the y-axis with the graph of the function, in which case it is (0,−1/9) as you said. It can also mean just the y-coordinate of that point, which is −1/9.Your reasoning about the x-intercepts is also basically correct. You have shown that there is no x-intercept. That happens for many functions, so don&#039;t worry about this function being an exception. A function may have no x-intercept, one, two, many, countable infinitely many, uncountable infinitely many.A function can have only one or no y-intercepts, however, due to the definition of a function.

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