If the graph of an equation intersects the x-axis, is it possible for there to be a horizontal tangent I would add a picture of the equation that this question pertained to, but the file size is too large
The equation is $x^2 + 2x + y^4 + 4y = 5$. 

The question was "Is it possible for this curve to have a horizontal tangent at points where it intersects the x-axis?" 
However, if it intersects the x-axis, then how could it have a horizontal tangent at the x-axis? To have a horizontal tangent at the x-axis, wouldn't the graph have to have some local minimum or maximum at the x-axis, and thus the point would actually be tangent to the x-axis, not intersecting it?
So I argued something along those lines and , apparently, I was wrong; the teacher gave the example of y = $x^2$, and said that it intersects the x-axis and has a horizontal tangent at the x-axis. I'm pretty sure $y = x^2$ is tangent to the x-axis though??? I'm confused; maybe I am assuming the wrong definitions for intersect and tangent. Can someone please provide insight?
Edit : After considering $y=x^3$ I realized that it is possible for a curve to cross the x-axis at a point and be tangent to it at that point. I wonder, though, if this is a unique property of points of inflection or if there is another reason. Also, I realized that intersect may not necessarily mean that the equation crosses at the x-axis; it could instead simply touch it. 

I guess it depends on your definition of "intersection", but my understanding is it means the two lines "touch" each other at least one point, regardless of whether or not these lines cross each other.
Note Intersection - math word definition states

An intersection is a single point where two lines meet or cross each other.

with this agreeing with my understanding. Thus, this Web page also says there's the option to just "meet", like what $f(x) = x^2$ does at $x = 0$, where it touches the $x$-axis. In addition, it's a local minimum since $f'(x) = 2x = 0$ there and $f(x) = x^2 \ge 0$ for all real $x$. Finally, since $f'(x) = 0$ there, the tangent to $f(x)$ is also a horizontal line.
