Hello Am I right in saying: ax+b=0 is one-variable linear equation ax+by+c=0 is two-variable linear equation ax^2+bx+c=0 is one-variable quadratic equation ax^2+bx+c=y is two-variable quadratic equation Every linear or quadratic equation in one or two variables can be represented in those ways. How come graph of ax+by+c=0 is point of solution set {(x,y) | ax+by+c=0}? Why not (y,x)? Since we can rearrange it as x=(y=b)/m, can we pick the y value first and then find the x value? I know graph of x=a and y=a is vertical and horizontal lines respectively because x or y will always be constant no matter what y or x is respectively. But how would we represent the solution sets? For x=a, would it be {(x,y) | x+0y=a}? Thanks.
Sure. I would say the general two-variable quadratic equation has the form [tex]ax^2 + by^2 + cxy + dx + ey + f = 0[/tex] It doesn't matter much. You might as well choose [tex]\{(y,x)~\vert~ax+ by + c = 0\}[/tex] It might not be the same set, but it will be an equivalent situation you're dealing with. Geometrically, this corresponds to just swapping X-axis and Y-axis. Usually, we will of course use [tex]\{(x,y)~\vert~ax + by + c = 0\}[/tex] but that's a convention. Yes, that will be one possible way of writing the solution set. Another one is [tex]\{(x,y)\in \mathbb{R}^2~\vert~ x =a\}[/tex] or [tex]\{(a,y)\in \mathbb{R}^2~\vert~y\in \mathbb{R}^2\}[/tex]
Am I right in saying: graph of f(x) is graph of solution set {(x,f(x)) | f(x)=x+a} Can I right it as x=f(x)-a too?
I would say the graph of a function ##f:\mathbb{R}\rightarrow\mathbb{R}## is [tex]\{(x,f(x))\in \mathbb{R}^2~\vert~x\in \mathbb{R}\}[/tex] or perhaps [tex]\{(x,y)\in \mathbb{R}^2~\vert~y=f(x)\}[/tex] In the case that ##f(x) = x+a## for all ##x##, then this becomes [tex]\{(x,x+a)\in \mathbb{R}^2~\vert~x\in \mathbb{R}\}[/tex] or [tex]\{(x,y)\in \mathbb{R}^2~\vert~y=x+a\}[/tex] and you can write this of course as [tex]\{(x,y)\in \mathbb{R}^2~\vert~x=y-a\}[/tex]
If your ##f## is defined by ##f(x) = x+a##, then that just says [tex]\{(x,f(x))~\vert~x+a=x+a\}[/tex] which simplifies to [tex]\{(x,f(x))~\vert~0=0\}[/tex] So yes, it's not really right.