I'm not the best at algebra so I will try to word this in the best way I can.
Let's say I have 2 solutions in each of 2 solution sets:
Set 1: (6, 0) and (18, 0) (visual here)
Set 2: (0, 0) and (2, 0) (visual here)
As you can see, the y position of the vertices for both of the equations shown above are different (set 1's vertex has a y position of about -36 and set 2's vertex has a y position of about -1)
How would I go about writing a quadratic equation that incorporates 2 solutions (for example 6 and 18 as shown in set 1) and makes the y position of the vertex always be at y = 10? Again I'm looking for an equation that will do this for any 2 solutions.
If you would like more clarification, ask and I will try to provide some more information. Any help is appreciated!
Let's define some terms that will help with the question.
The roots or zeros of a polynomial of degree $n$ are the solutions to the polynomial. On a graph, they are shown as the $x$-intercepts.
The 'vertex' of a quadratic graph is called a turning point, which can be minimum or maximum. It is the point which the gradient is equal to zero.
A useful form of writing a quadratic is called the turning point form, which is what you need at the moment. Turning point form is said to be written in the form:
$$y = a(x-h)^2+k$$
where $a$ measures the width of the parabola, and $(h,k)$ is the turning point of the graph.
If you are looking for a quadratic which has its turning point at $y=10$, then any quadratic will be in the form
$$y = a(x-h)^2 + 10$$
Any real value of $h$ will work, but $a$ must be negative - this reflects the graph along the x-axis. By doing this, the graph is upside-down (we call this concave down) and there will always be two solutions.
Therefore, to complete your problem, a quadratic in the form $y=a(x-h)^2+k$ such that $a<0$, $k=10$ will give a turning point at $y=10$ with two distinct solutions.
You can always expand the brackets to get to the normal quadratic form $y = ax^2 + bx + c$.
Let's do an example: let $a=-2$ and $h=3$ and $k=10$.
then $$y=-2(x-3)^2+10$$
the graph of $-2x^2+12x-8$ looks like this.