If I have a quadratic in the form of $f(x) =ax^2+bx+c$, how would I be able to transfer this onto a 3D Graph where $x$=horizontal, $y$=the other 2D dimension and $z$= height , and raise the turning point by $k$ in the $z$ axis, how might I find the equation of this 3D Graph$?$
Diagram (1) visualisation You could imagine the vertex of the green curve (the parabola), being shifted up/down the red vertical line by $k$ units, while the $x$ intercepts stay still. And then that is the new function I need the equation for.
Diagram (2) angle calculations
In that picture, I have made the quadratic: $f(x)=-x^2+10x$, and I want to shift it up by $1$ unit in the $z$ axis. I have drawn what that new function would look like with a red pen. I am keeping the $x$ intercepts the same, except since I am pulling the vertex upwards, the distance $AB>AC$, (look carefully at the diagram to see the letters A B and C on the graph), so the quadratic would have stretched upwards. You could imagine that $BC$ is part of the line that I am shifting the vertex upon.
I let $k=\theta=\angle BAC$, and draw a triangle $\triangle ABC$ onto the 3D graph, where $A$ is at the $x$ coordinate of the turning point of $f(x)$ which is $x=-b/2a=-10/-2=5$
Only the $x$ value though so I put it on the $x$ axis at $x=5\to (5,0,0)$
I make point $B$ at the turning point of my lifted curve $(x,y,z)\to(5,25,1)$
I make the point $C$ at the vertex of the original function $f(x)$ which is at $(5,25,0)$.
It is quite simple now to draw the triangle seperate from the graph. All lengths from here on are absolute and don't factor in angles. Distance $AC$ = $25$ units, $BC$=$1$ unit, and $AB = \sqrt{AB^2+BC^2} = \sqrt{626}$ units. And by using basic trigonometry, $\theta=\angle BAC \approx 2.29°$
How do I get my new function?
So the equation of your curve, before transformation, is this: $$ y = ax^2 + bx + c \\ z = 0 $$ And you'd like to transform it so that the vertex of the parabola lies at some $z$-value. Let's simplify a little, and rewrite the parabola in a vertex form: $$ y = (x-u)^2 + w\\ z = 0 $$ which has vertex at $(u, w, 0)$. You have a bit of a problem in your goal in the case where $w = 0$, for now the vertex is an intercept, so "raising the vertex" and "the intercepts stay still" is a contradiction.
Let me proceed with the case where $w \ne 0$. Then the intercepts occur when $y = 0$, i.e., when \begin{align} (x-u)^2 +w &= 0 \\ (x-u)^2 &= -w \\ (x-u) &= \pm\sqrt{-w} \\ x &= u \pm \sqrt{-w} \end{align}
So in the case where $w < 0$, this has solutions; for $w \ge 0$, your question doesn't even make sense (i.e., either the vertex is the same as the intercepts ($w = 0$) or there are no intercepts at all $(w > 0)$.
Continuing with $w < 0$, we have intercepts at $$ (u - \sqrt{-w}, 0, 0)\\ (u + \sqrt{-w}, 0, 0) $$ and you'd like a "shear" operation (a linear transformation where $x$ and $y$ stay fixed, but $z$ is altered) in which these points remain fixed. So we're looking for a map $$ (x, y, z) \mapsto (x, y, z + px + qy) $$ where the $y = 0$ line remains fixed, so $$ (x, 0, z) \mapsto (x, 0, z + px) = (x, 0, z) $$ hence $p = 0$.
And we also want to have $$ (u, w, 0) \mapsto (u, w, H) $$ for some target height $H$. So we need $$ (u, w, 0) \mapsto (u, w, 0 + qw) = (u, w, H) $$ hence $qw = H$, so $q = H/w$. Whew! Now we've got all we need. We need to transform everything by the map $$ (x, y, z) \mapsto (x, y, z + \frac{H}{w} y) $$ The "before" equation was \begin{align} y &= (x-u)^2 + w\\ z &= 0 \end{align}
The "after" equation is therefore
\begin{align} y &= (x-u)^2 + w\\ z &= \frac{H}{w}\left((x-u)^2 + w \right) \end{align}
Let's make this concrete with an example. Suppose we have $$ y = x^2 + 10x $$ which we rewrite in vertex form as $$ y = (x+5)^2 - 15 $$ so that the vertex is at $(u, w) = (-5, -15)$.
We want to "lift" the vertex up three units in $z$ (I just picked $3$ arbitrarily). So our formula says to plot \begin{align} y &= (x+5)^2 - 15\\ z &= \frac{3}{-15}\left( (x+5)^2 - 15 \right) \end{align}
Here's the plotted result, using geogebra: