# Why can I not make this function equal zero and solve for x when finding y-intercept of a rational fraction w/ numerator (2x^2+9)

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?

• 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 $y=0$. – Jaap Scherphuis Dec 2 '20 at 12:32

It would be $$x^2=\color{red}-\dfrac92$$, which has no real solutions.
You wrote $$x^2 = \frac{9}{2}$$. But this is false. It should read $$x^2 = -\frac{9}{2}$$.