How is it possible to have a graph of $e^y $ against $x^2$ that crosses the origin?

Question

I found this question from a Facebook group for high school IB students to ask about math questions. The asker asked a question that I don't seem to understand as well, how is it possible for a graph of $e^y$ against $x^2$ to ever pass through the origin as $e^y$ is always $>0$ and thus can never be zero. I'm wondering if there is some loophole in my understanding of the meaning 'a graph of A against B'?

From my understanding of the 'graph of $e^y$ against $x^2$, the $y$-axis of the graph will be $e^y$ and the $x$-axis will be $x^2$. This is different from e.g. when we plot a graph of the function $f(x)=x^2$ in which the $y$-axis will be $x^2$ and the $x$-axis will be $x$.

Furthermore, the equation format of the question will be $e^y=4(x)^2$ and thus this suggests that if $x^2=0$ then $e^y=0$ which I'm not sure if this argument holds true.

Is it possible that the question is proposing a theoretical case? Or does the theory that underly the question's case (which one does not need to know to do this question) involves imaginary numbers?

Answer 1

Hint $y$ is a function of $x$. Try to plot $e^{y(x)}$ versus $x$ when $y(x)=\ln x$

Answer 2

Well, $\displaystyle\lim_{y\to -\infty} e^y = 0$, so for huge negative values of $y$, the value $e^y$ is pretty close to zero, i.e., saying that $e^y=0$ (which is not mathematically correct but serves our purpose) means that $y$ is near negative infinity.

On a side note, since you noted that $e^y>0$, that means the "straight line" described in (d) is actually restricted to the first quadrant, i.e., it doesn't spawn across the whole plane (although including the origin as part of the "straight line" may be an abusive interpretation of the limiting case $\displaystyle\lim_{y\to -\infty} e^y = 0$.)