Crazy Equation

Well, technically, almost a line. At $x=0$ , it's indeterminate.

Everything cancels out so that the equation becomes $y=1$ .

First: $sgn(x)=\frac{\vert x\vert}{x}$ so the denominator cancels.

$y=sin^2(\dfrac{x\pi}{7})*ln(e)+cos^2(\dfrac{x\pi}{7})$

Second: ln(e)=1

$y=sin^2(\dfrac{x\pi}{7})+cos^2(\dfrac{x\pi}{7})$

Third: Sin^2(x)+Cos^2(x)=1

$\therefore \boxed{y=1}$

$\textit{trolololololololololololololo}$

sgn(x) = |x|/x by definition so these two terms cancel out.

ln(e) = 1

and sin^2(x) + cos^2(x) = 1

So we have a line y = 1

oh WOW I was fooled by its looks and ran to WolframAlpha... But I do gotta admit I had no idea what sgn(x) was, which is probably why more than anything.

But yeah it would be ALMOST a line, but since the question asks what it would LOOK LIKE not what it would BE, then the answer choice is correct.