In a quadratic function $(f(x) = ax^2+bx+c)$, are $a$, $b$, and $c$ all limited to real numbers? If not, can $b = i$? ex: $(f(x) = ax^2 + ix + c)$? How would this affect the graph and quadratic formula to find its $x$-intercepts.
$x$ is still a real number.
Range of $x$: $(-∞,∞)$.
$a$, $b$, and $c$ are constants.
Also, I am sorry about the poor formatting earlier which had lead to misinterpretations.
On
For real $x$,
$$ax^2+ix+c$$ represents a complex number with real part $ax^2+c$ and imaginary part $x$.
In the complex plane, you can see it as a parabola of parametric equation
$$\begin{cases}u=ax^2+c,\\v=x\end{cases}$$
or after elimination of $x$,
$$u=av^2+c.$$
You can picture this in 3D, using the coordinate system $(x,u,v)$. It still gives a parabola, inscribed in the plane $v=x$ (the first bissector plane of $xu$ and $uv$).
If you keep calling the $x$-intercepts the intersection points of the axis $x$ with the curve, there aren't any. Because $u=v=0$ is $ax^2+c=x=0$, which is impossible (unless $c=0$).
More generally, all coefficients can be complex, so that $ax^2+bx+c$ is
$$\begin{cases}u=a_rx^2+b_rx+c_r,\\v=a_ix^2+b_ix+c_i.\end{cases}$$
By combining these two equations you can obtain a relation of the form
$$x=pu+qv+r$$
so that
$$u=a_r(pu+qv+r)^2+b_r(pu+qv+r)+c_r,$$ still a parabola (in the complex plane).
If $x,a,b,c$ are real numbers (purely real complex) then the $f(x) = ax^2 + bx + c$ is a function from $\mathbb R\to \mathbb R$. The graph is one dimensional curve existing in two-dimensional space.
If $x$ is real and $a,b,c$ are (not all purely real) complex numbers then $f(x)$ is a function from $\mathbb R\to \mathbb C$. The graph is a one dimensional curve existing in three-dimensional space.
[Edit: After considering Yves Daoust's answer and observation, I should point out that a function: $\mathbb R \to \mathbb C$ via $f(x) = (a_r + a_i i)x^2 + (b_r + b_ii)x + (c_r + c_i i) = (a_rx^2 + b_rx +c_r) + (a_ix^2 + b_ix + c_i)i$ the graph of which, does exist in 3-D space, it does so as a parabola contained in a plane; just not one of the planes defined by two of the dimensional axes.]
If $x$ is complex (regardless of whether $a,b,c$ are all purely real complex or not) then $f(x)$ is a function from $\mathbb C \to \mathbb C$. The graph is a two-dimensional surface in a four-dimensional space.
The quadratic equation that $a^2z + bz + c = 0$ will have solutions $z = \frac {-b \pm \sqrt {b^2- 4ac}}{2a}$ will hold true whether $a,b,c$ or $z$ are are real or complex numbers.
So if $b = i$ and the function is $f(x) = ax^2 + ix + c$ then we can solve $f(x)=0$ via $x = \frac {-i\pm\sqrt {-1 - 4ac}}{2a}$.
(Note: if $x$ is restricted to be purely real, then this value, $\frac {-i\pm\sqrt {-1 - 4ac}}{2a}$ may not be purely real and not be a legitimate value for $x$. In fact, if $a$ is real, then this can only be purely real if $-1 + 4ac = (k+i)^2$ for some real $k$ and then only $\frac {-i +(k+i)}{2a}=\frac k{2a}$ will be a solution; $\frac {-i-(k+i)}{2a} = \frac {-k-2i}{2a}$ will not be an acceptable real value for $x$.)
(Hmm, and if $c$ is real then $k = 0$ and $4ac=0$ and $c = 0$.... but if $a,$ can be non-real... well if $x$ must be real we still must have a purely real value for this to have a solution.)
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This assumes that $a, b,$ and $c$ are constants.
You can not replace $b$ (or $a$ or $c$) with a value dependent upon $x$.
I first misread your question as replacing $b$ with $i(ax^2+ix + c)$. Although this seems not to have been your intent my initial explanation still is worth observing:
If we replace $b$ with $i(ax^2+ix + c)$, we would have the function $f(x) = (a+i)x^2 + ix + (c+i)$. This is a different quadratic.
And you can use the quadratic equation to solve it for $f(x) = 0$ then $x=\frac {-i^2 \pm \sqrt {i^2 + 4(a+i)(c+i)}}{2(a+i)}$. If $x$ is restricted to real values and the graph $(x,f(x))$ is a one dimensional curve in three dimensional space, these values may not be real, and there may be no real solution.