= Solution
The <fundamental theorem of calculus> gives $y'(x)=e^{-x^2}>0$, so the graph is strictly increasing. Its second <derivative> is $y''(x)=-2xe^{-x^2}$: it curves upward for $x<0$ and downward for $x>0$, with an <inflection point> at zero.
By the <Gaussian integral> and the evenness of its <integrand>,
$$
\boxed{y(0)=\frac{\sqrt\pi}{2}},\qquad
\lim_{x\to-\infty}y(x)=0,\qquad
\lim_{x\to+\infty}y(x)=\sqrt\pi.
$$
Thus the only axis intersection is $(0,\sqrt\pi/2)$; there is no finite $x$-axis intersection because the <integral> is strictly positive for every finite $x$. The <error function> gives $y(x)=\tfrac{\sqrt\pi}{2}(1+\operatorname{erf}x)$.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-2-q5-gaussian-integral.png]
{title=The increasing integral of a Gaussian}
{description=The graph has its only axis intersection and its <inflection point> at $(0,\sqrt\pi/2)$. The horizontal lines at zero and $\sqrt\pi$ are asymptotes.}
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