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When graphing a rational fraction, $\frac{2x^2+9}{x}$, and I have to find the $y$-intercept, why can't I do this? Prof. Leonard from youtube says that if discriminant is negative, then there is no real answer, but why do I get a value for $x$ when I do this?

$$2x^2+9 = 0 \to 2x^2= -9 \to x^2 = \frac{9}{2} \to x = \pm\frac{3}{\sqrt{2}}$$

$$x = \frac{-3}{\sqrt2}, x=\frac{3}{\sqrt2}$$

What am I doing wrong?

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    $\begingroup$ The $y$-intercept is where the graph intersects the $y$ axis, so where $x=0$. This function is not defined at $0$ (and actually has an asymptote there). You seem to be trying to find the $x$-intercept(s), the points where the function intersects the $x$-axis, i.e. points for which $y=0$. $\endgroup$ – Jaap Scherphuis Dec 2 '20 at 12:32
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What am I doing wrong?

You mistakenly dropped a minus sign.

It would be $x^2=\color{red}-\dfrac92$, which has no real solutions.

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  • $\begingroup$ Yes I see it now, thank you! $\endgroup$ – Science learner Dec 2 '20 at 11:55
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You wrote $x^2 = \frac{9}{2}$. But this is false. It should read $x^2 = -\frac{9}{2}$.

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