Given the following function: $$f(x)=ax^2 + bx$$where $a$ and $b$ are constants and $a,b\in \mathbb{R}$. The graph of $y$ against $x$ where $y=f(x)$ will be a quadratic. But I can convert the equation $f(x)=ax^2+bx$ into its linear form, $Y=mX+c$, so that the graph of the new equation will be a linear graph.
This can be done by dividing the LHS and RHS of the equation by $x$, giving me, $$\frac{f(x)}{x}=\frac{ax^2+bx}{x} = ax+b$$
Graphing $Y$ against $X$, where $Y=\frac{f(x)}{x}$ and $X=x$, will give me a linear graph. The slope of the line will be $a$ and the $Y$-intercept of the line will be $b$. Except, I'm not sure if the $Y$-intercept does exist. The $Y$-intercept of the line can be obtained from the line equation when $X=0$. But doing so gives me
$$Y=\frac{f(0)}{0}=a(0)^2+b$$
From my knowledge, I think that there should be a hole on the line when $X=0$ since it will lead to division by $0$ which is not defined. Hence, the $Y$-intercept of the line doesn't exist. But my teacher disagree, saying that the $Y$-intercept exist but I can't solve for $f(x)$ when $X=0$ ($Y=b$ when $X=0$ but $f(x)$ is not solvable). So I'm a little confused if the $Y$-intercept of this line exist. Maybe I miss something that allow this to happen. I would appreciate any help. Thanks.
It is not quite the case that
$$\frac{ax^2+bx}{x} = ax+b.$$
Whenever you divide by a variable quantity, you have to say, "either this quantity is zero or I can divide by it." So to be precise, you'd have to write
$$\frac{ax^2+bx}{x} = ax+b, x\neq 0.$$
So the $y$-intercept doesn't exist. There is a hole there, as you expected.